2
Functions of a Complex Variable
2.1 Introduction
Many engineering problems can be solved using Complex Variable Theory. Some of the elementary problems dealing with applications to electric circuits or mechanical vibrating systems need only elementary knowledge of complex analysis. Advanced problems in heat conduction, fluid flow and electrostatics require knowledge of complex analysis.
Equations such as x^{2} = –1 have no solutions in real variables. By defining i = as a complex unit, the system of real numbers can be extended to the complex number system.
2.2 Complex NumbersComplex Plane
A complex number z is an ordered pair of real numbers x and y written as z = (x, y). Here x is called the real part and y the imaginary part of z. We write
Equality of two complex numbers
Two complex numbers z_{1} = (x_{1}, y_{1}) and z_{2} = (x_{2}, y_{2}) are equal if and only if their real parts are equal and their imaginary parts are equal, i.e.,
Imaginary unit
(0, 1) is called the imaginary unit and is denoted by i.
Addition and multiplication
Addition of two complex numbers z_{1} = (x_{1}, y_{1}) and z_{2} = (x_{2}, y_{2}) is defined by
z_{1} + z_{2} 
= 
(x_{1}, y_{1}) + (x_{2}, y_{2}) 


= 
(x_{1} + x_{2}, y_{1} + y_{2}) 
(2.2) 
Multiplication is defined by
z_{1} · z_{2} 
= 
(x_{1},y_{1})(x_{2},y_{2}) 


= 
(x_{1}x_{2} − y_{1}y_{2}, x_{1}y_{2} + x_{2}y_{1}) 
(2.3) 
In particular, these two definitions imply that
as for real numbers x_{1} and x_{2}. Hence the complex numbers extend the reals. We can write
(x, 0) 
= 
x, (0,y) = iy, (x,y ∈ ℝ) 
(2.4) 
∵ iy 
= 
(0, 1)(y, 0) 


= 
(0 · y − 1 · 0,0 · 0 + 1 · y) 


= 
(0, y). 

Since (x, y) = (x, 0) + (0, y) = x + iy it has great practical advantage to write z = x + iy. If x = 0 then z = iy and is called a pure imaginary number. Also Eqs. (2.1) and (2.3) give
By the standard notation, addition may be done as follows:
Multiplication can be done in the usual way noting that i^{2} = –1
Real and imaginary parts
Let z_{1} = 3 – 4i; z_{2} = –5 + i.
Then Re z_{1} = 3 Im z_{1} = –4
Re z_{2} = –5 Im z_{2} = 1
Sum and product
z_{1} + z_{2} 
= 
(3 − 4i) + (−5 + i) = −2 − 3i 
z_{1}z_{2} 
= 
(3 − 4i)(−5 + i) 

= 
(−15 + 4) + i(3 + 20) = −11 + 23i 
Subtraction and division
Subtraction and division are inverse operations of addition and multiplication.
Let z_{1} = (x_{1}, y_{1}) and z_{2} = (x_{2}, y_{2}). The difference z_{1} – z_{2} is the complex number z = (x, y) such that
z + z_{2} = z_{1} 
⇒ 
(x,y) + (x_{2},y_{2}) = (x_{1},y_{1}) 

⇒ 
(x + x_{2},y + y_{2}) = (x_{1},y_{1}) 

⇒ 
x + x_{2} = x_{1}, y + y_{2} = y_{1} 

⇒ 
(x,y) = (x_{1} − x_{2},y_{1} −y_{2}). 
(x + iy) + (x_{2} + iy_{2}) 
= 
(x_{1} + iy_{1}) 
⇒ x + x_{2} 
= 
x_{1}, y + y_{2} = y_{1} 
⇒ x 
= 
x_{1} − x_{2}, y = y_{1} − y_{2} 
⇒ z 
= 
z_{1} − z_{2} 

= 
(x_{1} − x_{2}) + i(y_{1} − y_{2}). 
The quotient z_{1} / z_{2} (z_{2} ≠ 0) is the complex number z such that zz_{2} = z_{1}
(x + iy)(x_{2} + iy_{2}) 
= 
(x_{1} + iy_{1}) 
⇒ xx_{2} − yy_{2} 
= 
x_{1}, xy_{2} + yx_{2} = y_{1} 
xx_{2} − yy_{2} 
= 
x_{1} ∣x_{2} ∣y_{2} 
xy_{2} + yx_{2} 
= 
y_{1} ∣y_{2} ∣ x_{2} 

= 
x_{1}x_{2} + y_{1}y_{2} 

= 
x_{2}y_{1} − x_{1}y_{2} 
Solving z = (x, y)
In practice, we do as follows:
Figure 2.1
Difference and quotient
Let z_{1} = 3 − 2i and z_{2} = 4 + i
z = z_{1} − z_{2} = (3 − 2i) − (4 + i) = −1 − 3i
Complex numbers satisfy
laws w.r.t the operations of addition and multiplication and also the distributive laws as real numbers and form a field.
ℚ, ℝ, ℂ are fields; ℚ,ℝ satisfy order relation ‘<’; ℝ is complete ordered field. No such order relation exists in ℂ.
There is oneone correspondence between the set of reals ℝ and points in a line, and the set of complex numbers ℂ and the points in a plane.
Geometrical Representation of Complex Numbers
Complex plane or Argand plane
Choose two perpendicular lines, the horizontal line called the real axis and the vertical line called the imaginary axis. Thus we have a cartesian coordinate system. Plot a given complex number z = (x,y) = x + iy as a point P with coordinates x and y.
Addition and subtraction can be visualized as shown above (Fig. 2.1)
Complex conjugate numbers
The complex conjugate of a complex number z = x + iy is defined by
Replace i by –i in the complex number to get its conjugate. It is obtained geometrically by reflecting the point z in the real axis.
Figure 2.2
We have z = x^{2} + y^{2}; z + = 2x; z − = 2iy
If z is real, z = x, then = z. Conversely if = z then z is real and z = x.

Conjugate of sum = Sum of conjugates

Conjugate of difference = Difference of conjugates

Conjugate of product = Product of conjugates

Conjugate of quotient = Quotient of conjugates
Example 2.1
Let z_{1} = 2 − 3i, z_{2} = 4 + 5i
EXERCISE 2.1
 Powers of the imaginary unit:
Show that i^{2} = −1, i^{3} = −i, i^{4} = 1, i^{5} = i, ⋯ and
 Multiplication by i is geometrically a counterclockwise rotation through π/2 (90°). Verify this by plotting z and iz and the angle of rotation for z = 4 + 2i, z = –1 + i, z = 5 – 2i.
Figure 2.3
 Let z_{1} = –2 + 3i, z_{2} = 1 + i. Find
 z_{1}z_{2} = (–2 + 3i)(1 + i)
= (−2 − 3) + i(3 − 2) = −5 + i
 Re(z_{1})^{2} = Re(5 – 12i) = 5 (Re z_{1})^{2} = (–2)^{2} = 4 z = x + iy
 (1 + i)^{2} = 2i; Re(1 + i)^{2} = 0; (1 + i)^{4} = (2i)^{2} = −4
 z_{1}z_{2} = (–2 + 3i)(1 + i)
Polar Form of Complex Numbers, Powers and Roots
If we put
x = r cos θ, y = r sin θ
z = x + iy takes the polar form
z = r(cos θ + i sin θ) or re^{iθ}.
r is called the absolute value or modulus of z and is denoted by ∣z∣. ∣z∣ = r =
Figure 2.4
Geometrically, ∣z∣ represents the distance of the point from the origin.
The inequality ∣z_{1}∣ > ∣z_{2}∣ means that the point z_{1} is farther away from the origin than the point z_{2}, and ∣z_{1} − z_{2}∣ is the distance between the points z_{1} and z_{2}. θ is called the argument or amplitude of z denoted by arg z or amp z.
Geometrically, θ is the directed angle from the positive xaxis to OP. For z = 0, this angle θ is undefined. For z ≠ 0 it is determined only up to an integer multiple of 2π. The value of θ that lies in the interval –π < θ ≤ π is called the principal value of the argument of z ≠ 0 and is denoted by Arg z with capital A. Thus –π < Arg z ≤ π.
Example 2.2
Arg z = (principal value).
Since for any complex number z = x + iy,
Figure 2.5
Similarly z ≥ y
∴ Re z ≤ z; Im z ≤ z.
Let z_{1} and z_{2} be complex numbers. Then the origin and the points z_{1} and z_{1} + z_{2} are the vertices of a triangle whose sides are z_{1}, z_{2} and z_{1} + z_{2}. We obtain triangle inequality
Also, we have
(The difference of two sides of a triangle cannot exceed the third side – Fig. 2.6.)
Figure 2.6
Multiplication and Division in Polar Form
Let z_{1} 
= 
r_{1} (cos θ_{1} + i sin θ_{1}), 
z_{2} 
= 
r_{2} (cos θ_{2} + i sin θ_{2}) 
z_{1}z_{2} 
= 
r_{1}r_{2} [(cos θ_{1} cos θ_{2} − sin θ_{1} sin θ_{2}) +i(sin θ_{1} cos θ_{2} + cos θ_{1} sin θ_{2})] 

= 
r_{1}r_{2}cis (θ_{1} + θ_{2}) 
∣z_{1}z_{2}∣ 
= 
∣z_{1} ∣∣z_{2}∣ 
arg (z_{1}z_{2}) = arg z_{1} + arg z_{2}, up to multiple of 2π
arg (zz_{2}) = arg z + arg z_{2} = arg z_{1}
arg = arg z_{1} − arg z_{2}, up to multiple of 2π
Example 2.3
Division
The quotient z = z_{1} /z_{2} is the number z satisfying zz_{2} = z_{1}.
Hence zz_{2} = zz_{2} = z_{1}. arg zz_{2} = arg z + arg z_{2} = arg z_{1}. This gives z_{1}/z_{2}=z_{1}/z_{2} (z_{2} ≠ = 0) and arg(z_{1}/z_{2}) = arg z_{1} − arg z_{2}.
Integer Powers: DeMoivre’s Formula
DeMoivre’s formula
(cos θ + i sin θ)^{n} = cos nθ + i sin nθ
Put n = 2
cos^{2} θ − sin^{2} θ + 2i sin θ cos θ = cos 2θ + i sin 2θ
⇒ cos 2θ = cos^{2} θ − sin^{2} θ and sin 2θ = 2 sin θcos θ
Roots
z = w^{n} (n = 1,2,3,…) nth root of z
Remark
This symbol is multivalued (nvalued) in contrast to the usual convention in real calculus.
By equating absolute values root is real positive and unique.
By equating arguments Nφ = θ + 2kπ
thus
For these values of k, we get n distinct values of w; these values repeat themselves for higher values of k.
Consequently, for z ≠ 0 has the n district values
where k = 0, 1, … (n – 1). These n values, lie on a circle of radius with centre at the origin and constitute the vertices of a regular polygon of n sides.
The value of obtained by taking the principal value of Arg z and k = 0 in Eq. (2.6) is called the principal value of w = In particular, taking z = 1 we have z = r = 1 and Arg z = 0. Then Eq. (2.6) gives
These n values are called the nth roots of unity. They lie on the circle of radius 1 and centre 0, briefly called the unit circle (see Fig. 2.7).
If w denotes the value corresponding to k = 1 in Eq. (2.7) the n values of can be written as
Similarly if w_{1} is any nth root of an arbitrary complex number z then the n values of in Eq. (2.6) are
because multiplying w_{1} by w^{k} corresponds to increasing the argument of w_{1} by 2kπ∕n.
Figure 2.7
Definitions of Certain Terms
Distance between two points
If ‘z’ and ‘a’ are points in the complex plane ℂ the distance between them is d(z, a) = z − a
Set of points
By a set of points we mean a collection of points in the complex plane ℂ.
Example 2.4
Circle : S_{1} = {z ∈ ℂ  iz − a = ρ} is called a circle with centre ‘a’ and radius. ρ.
Right halfplane
S_{2} = {z ∈ ℂRe z > 0} is the right halfplane.
Neighbourhood of a jroint z_{0}
The set of all points z for which z − z_{0} < δ where δ is a positive constant, is called a neighbourhood of z_{0} and written nbd of z_{0}.
Deleted nbd of z_{0}
The set of all points z for which 0 < z − z_{0} < δ is called a deleted nbd of z_{0}.
Open circular disc
The set of all z for which z − z_{0} < ρ is called an open circular disc.
Closed circular disc
The set of all z for which z − z_{0} ⩽ ρ is called a closed circular disc.
Open annulus or circular ring
The set of all points z such that ρ_{1} < z − z_{0} < ρ_{2} (ρ_{1} < ρ_{2}) is called an open annulus.
Closed annulus or circular ring
The set of all points z such that ρ ⩽ z − z_{0} ⩽ ρ_{2} (ρ_{1} < ρ_{2}) is called a closed annulus.
Interior point of a set S
A point z_{0} is called an interior point of a set S if there is a nbd of z_{0} containing only the points of S.
Open set
A set S is called open if every point of S has a nbd every point of which belongs to S.
Example 2.5
 The points inside a circle.
 The points inside a square.
 Right halfplane Re z = x > 0.
Open halfplane
Open upper halfplane: The set of points z such that Im z = y > 0.
Open lower halfplane: The set of points z such that Im z = y < 0.
Closed set
A set S is said to be closed if its complement S’ is open.
Example 2.6
The set of points on and inside the unit circle z = 1.
Bounded set
A set S is called bounded if all its points lie within a circle of sufficiently large radius.
Example 2.7
Points in a rectangle, ellipse or polygon.
Note:
Points on a straight line or in an infinite strip do not form a bounded set.
Connected Set
A set S in which any pair of points can be joined by a polygonal arc entirely lying in S, is called a connected set.
Figure 2.8
Domain
An open connected set is called a domain.
Example 2.8
Interior of a circle or a square or a closed polygon.
Boundary point of a set S
A point every nbd of which contains both points in S and points not in S.
Example 2.9
Points on the two circles C_{1} and C_{2} of an annulus.
Note:
If S is open, no boundary point belongs to S. If S is closed, every boundary point belongs to S. z ⩽ 1: closed, bounded set.
Im z^{2} = xy > 0 : open, unbounded set.
ρ_{1} < z − z_{0} < ρ_{2}: open, bounded, unconnected set.
Region
A set of points consisting of a domain and some or all of its boundary points.
2.2.1 Complex Function
Let S be a set of complex numbers. A function f defined on S is a rule that assigns to every z in S a complex number w, called the value of f at z. We write
z varies in S and is called a complex variable and S is called the domain of definition of f.
Example 2.10
w = f (z) = z^{2} + 2z
Let z 
= 
x + iy; z^{2} = (x + iy)^{2} = x^{2} − y^{2} + 2ixy; 
2z 
= 
2x + i2y 
∴ w 
= 
u (x, y) + iv (x, y) = x^{2} −y^{2}+2x+i (2xy+2y) 
Re w 
= 
u(x,y) = x^{2} − y^{2} + 2x; Im w = 2xy + 2y 
2.2.2 Limit of a Function
Let f (z) be a function defined in some deleted nbd of z_{0} and l be a complex number. If for every ∈> 0 there corresponds a δ > 0 such that
then we say that f (z) tends to l as z tends to z_{0}. We write f (z) = l.
2.2.3 Continuity at z_{0}
Let f (z) be defined in a nbd of z_{0}. If f (z) = f (z_{0}) then we say that f is continuous at z_{0}.
Figure 2.9
2.2.4 Differentiability
A complex function f (z) is said to be differentiable at a point ‘z_{0}’, if the limit
exists. The limit f’(z_{0}) is called the derivative off (z) at z_{0}. Putting Δz = z – z_{0} we can write the limit as follows:
Remark
In real calculus, there are two directions – rightside and leftside – to approach a point x_{0}. The exists ⇔
 rightside limit exists
 leftside limit exists
 both are equal.
But in the case of complex numbers the situation is different. One can approach a point z_{0} in the complex plane from any of the infinite number of directions. The above limit must exist and be the same when we approach z_{0} in any of these directions. Thus the condition of differentiation is stringent in complex variables. Rules of differentiation are the same as in real calculus. All the complexvalued functions whose realvalued counterparts are differentiable are differentiable.
It is important to note that many simple functions do not have derivatives at any point.
Example 2.11
is not differentiable at any point z.
Solution Let f (z) = = x – iy. Writing Δz = Δx + iΔy
If Δy = 0 this is + 1 and if Δx = 0 this is –1. Thus Eq. (1) approaches +1 along path I but –1 along path II. Hence the limit of Eq. (1) as Δz → 0 does not exist at any point z.
Figure 2.10
Example 2.12
Show that f (z) = Re z = x is not differentiable at any z.
Solution
Example 2.13
f (z) = z^{2} is not derivable at any z except at z = 0.
Solution
Case (i) Δy → 0, Δx → 0
Case (ii) Δx → 0, Δy → 0
f′(z) exists ⇔ z + = −z + ⇒ 2z = 0 ⇒ z = 0 f′(z) does not exist though f (z) = z^{2} is continuous at all z (for z ≠ 0).
2.2.5 Analytic Functions: Definition of Analyticity
A function f (z) is said to be analytic in a domain D if f (z) is defined and differentiable at all points of D.
The function f (z) is said to be analytic at a point z = z_{0} in D if f (z) is analytic in a nbd of z_{0} in D.
By an analytic function we mean a function that is analytic in some domain.
‘Regular’ and ‘holomorphic’ are other terms used for analytic functions. The fact that a function f (z) is holomorphic in a domain D is expressed as
2.2.6 ^{1}Cauchy–^{2}Riemann Equations
The Cauchy–Riemann equations (C.R.Es) provide us a test for the analyticity of a complex function
f is analytic in a domain D ⇔ (i) the first partial derivatives of u, v exist and are continuous in D, and (ii) u, v satisfy Cauchy–Riemann equations u_{x} = v_{y}, u_{y} = –v_{x} in D.
Example 2.14
f (z) = z^{2} = x^{2} − y^{2} + i2xy is analytic for all z
u = x^{2} − y^{2}, v = 2xy satisfy CREs u_{x} = 2x,
u_{y} = −2y, v_{x} = 2y, V_{y} = 2x ∴ u_{x} = v_{y} u_{y} = −v_{u}
2.2.7 Cauchy–Riemann Equations in Cartesian Coordinates
Theorem 2.1 If f (z) = u(x, y) + iv(x, y) is differentiable at z then at this point the first order partial derivatives of u and v exist and satisfy the CREs
Proof: By hypothesis f is differentiable at z. So
Hence the limit value along Paths I and II must be the same.
Figure 2.11
Choose Path I. We let Δy → 0 first and then Δx → 0. After Δy = 0, Δz = Δx, then Eq. (2.10) becomes
Similarly along Path II, we let Δx → 0 first and then Δy → 0. After Δx = 0, Δz = iΔy, then Eq. (2.10) becomes
∴ Existence of f^{′}(z) implies the existence of u_{x}, v_{x}, u_{y}, v_{y} in Eqs. (2.11) and (2.12).
Equating the real and imaginary parts of Eqs. (2.11) and (2.12) we get the CREs.
Theorem 2.2 If two realvalued functions u(x,y), v(x,y) of two real variables x and y have continuous first partial derivatives that satisfy CREs in some domain D then the complex function f (z) = u(x,y) + iv(x,y) is analytic in D.
Proof: It is given that are continuous.
We have,
where ∈_{1}→ 0, η_{1} → 0 as Δx → 0 and Δy → 0.
Again it is given that are continuous.
So we have,
where ∈_{2}→ 0, η_{2} → 0 as Δx → 0 and Δy → 0.
∴ Δw = Δu + iΔv
where ∈ = ∈_{1} + i ∈_{2} → 0, η = η_{1} + iη_{2} → 0 as Δx → 0 and Δy → 0.
Using CREs, we can write Eq. (2.15) as
Dividing both sides by Δz = Δx + iΔy and taking limit as Δz → 0
So, the derivative exists and is unique.
Alternative forms of CREs
Theorem 2.3 (Complex form of CREs). Let f (z) = u(x, y) + iv(x, y) be differentiable in a domain D. Then the CREs u_{x} = v_{y} and u_{y} = –v_{x} can be put in a complex form as
Proof: Let
f (z) 
= 
u(x,y) + iv(x,y). Then, 
f_{x} 
= 
u_{x} + iv_{x},f = u_{y} + iv_{y} 
Now, f_{x} 
= 
− if_{y} 


⇔ u_{x} + iv_{x} = −i(u_{y} + iv_{y}) 


⇔ u_{x} = v_{y} and u, = −v_{x} 
2.2.8 CauchyRiemann Equations in Polar Coordinates
Theorem 2.4 (CREs in polar coordinates). If f (z) = u(r, π) + iv(r, π) is differentiable at z = re^{iπ} then .
Also, .
Proof: We know that x = r cos θ, y = r sin θ
and
Similarly, we can prove that by
CREs.
Now,
CRES in Polar Coordinates [Alternative Method]
In polar coordinates, CREs are
Let x = r cos θ, y = r sin θ. Then z = x + iy = r cos θ + ir sin θ = re^{iθ}.
So, u + iv = f (z) = f (re^{iθ})
Differentiating partially w.r.t r and θ, we get
Equating the real and imaginary parts
Note:
 If f (z) = u(x, y) + iv(x,y) is analytic in a domain D then Re f = u = u(x,y) and Im f = v = v(x,y) have partial derivatives u_{x}, u_{y}, v_{x} and v_{y} and satisfy the partial differential equations u_{x} = v_{y} and u_{y} = –v_{x} called the Cauchy–Riemann Equations.
 The derivative f ′(z) can be calculated using either of the two formulas:
f′(z) = u_{x} + iv_{x} or f′(z) = −iu_{y} + v_{y}.
 Let f (z) = u + iv. Then the Cauchy–Riemann Equations u_{x} = v_{y} and u_{y} = –v_{x} are only necessary conditions for analyticity (differentiability) of f but not sufficient.
 Let f (z) = u + iv. Then a sufficient condition for analyticity of f is that u and v have continuous partial derivatives and also satisfy CREs u_{x} = v_{y} and u_{y} = –v_{x}.
Properties of analytic functions
 The sum, the difference, the product and the quotient of analytic functions are analytic. f, g ∈ H(D) ⇒f ±g,fg,f/g(g(z) ≠ 0) ∈ H(D).
 An analytic function of an analytic function is analytic.
 An entire function of an entire function is entire. (A function which is analytic in the entire complex plane is called an entire function, e.g. e^{z}, polynomial, sin z, cos z etc.)
 f ∈ H(D) ⇒ f is differentiable at all z ∈ D. ⇒ f is continuous at all z ∈ D.
 If f is analytic at a point z = a then f is differentiable at z = a. However, the converse is not true.
Example 2.15
We have proved that f (z) = z^{2} is differentiable only at z = 0 and nowhere else. This function is not analytic at z = 0 since according to the definition of analyticity at a point, f must be differentiable not only at z = 0 but at every point of some nbd of 0.
Example 2.16
If f (z) = z^{2} is analytic for all z, then the CREs are satisfied.
For f (z) = = x − iy we have u = x, v = −y
CRE(2) u_{y} = 0 = − v_{x} is satisfied
CRE(i) u_{x} = 1 ≠ v_{y} = − 1 is not satisfied.
∴ is not analytic.
An analytic function of constant absolute value is constant
Example 2.17
Show that if f (z) is analytic in D and f (z) = k, a constant in D then f (z) is constant in D.
Solution By assumption u^{2} + v^{2} = k^{2}.
By differentiating partially w.r.t x and y we get
uu_{x} + vv_{x} = 0, uu_{y} + vv_{y} = 0
⇒ (a) uu_{x} − vu_{y} = 0 (b) uu_{y} + vu_{x} = 0 (1)
∵ v_{x} = −u_{y}, v_{y} = u_{x} by CREs.
To eliminate u_{y}, multiply 1(a) by u and 1(b) by v and add. We get
Similarly to eliminate u_{x}, multiply 1(a) by –v and 1(b) by u and add. We get
If k^{2} = u^{2} + v^{2} = 0, then u = v = 0 ⇒ f = 0.
If k ≠ 0 then u_{x} = u_{y} = 0 ⇒ v_{x} = v_{y} = 0, by CREs.
∴ u = constant, v = constant ⇒ f (z) is constant of D.
2.2.9 MilneThomson’s Method
Example 2.18
Find the analytic function f (z) = u + iv where u(x,y) = e^{x} (x cos y – y sin y + cos y) + 2 cos x sinh y + 3x^{2} – 3y^{2}.
Solution Differentiating partially w.r.t x and y, we get,
u_{x} = e^{x}(x cos y − y sin y + cos y) + 2 cos x sinh y + 3x^{2} − 3y^{2}
u_{y} = e^{x} (− x sin y − sin y − y cos y) + 2 sin x cosh y − 6xy + 1
We know that f ′(z) = u_{x} + iv_{x} = u_{x} − iv_{y}.
Replace x by z and y by 0
∴ f′ (z) = e^{z}(z · 1 − 0) + e^{z} (1) + 0 + 3z^{2}−0−2i sin z−i
Integrating w.r.t z, we get
f (z) = 2e^{z} − e^{z} + e^{z} + z^{3} + 2i cos z − iz + c.
Limits, continuity, differentiability
Example 2.19
Find ‘a’ so that f is continuous.
Solution f (z) is continuous everywhere except at z = – i.
Path I :
Path II :
Thus = i = a = f (−i). Hence f is continuous if a = i, but not otherwise.
Example 2.20
Show that is not continuous at z=0.
Solution Let z = x + iy.
Path I :
Along the xaxis:
Path II :
Along the yaxis:
does not exist. Hence f is not continuous at z = 0.
Example 2.21
Show that f (z) = continuous at all z.
Solution We have f (z) − f (z_{0}) =  − _{0}. For any given ∈ choose ∈ = δ. We have z − z_{0} < δ ⇒ f (z) − f (z_{0}) <∈⇒ f (z) is continuous at z_{0}.
Since z_{0} is arbitrary f (z) is continuous for all z.
Example 2.22
Show that f (z) = z is continuous at all z.
Solution We have f (z) − f (z_{0}) = ⩽ z − z_{0}. For any ∈> 0 choose ∈ = δ then z − z_{0} < δ ⇒ f (z) − f (z_{0}) <∈⇒ f is continuous at z_{0}.
Since z_{0} is arbitrary f is continuous at all z.
Example 2.23
Show that does not exist though the function approaches the same limit along every straight line through the origin.
Solution
Path I :
Path II :
Path III :
Along any straight line y = mx through (0, 0),
Path IV :
Along the parabolic path y = mx^{2}
Along Path IV the limit is dependent on m.
Example 2.24
Solution
Path I :
Path III :
Since the limit does not exist f is discontinuous at z = 0.
Example 2.25
Show that every differentiable function is continuous.
Solution Let f (z) be differentiable at ‘a’. Then
∴ f (a) is welldefined. Consider
Hence f is continuous at z = a.
Remark
The converse of the above result is not true. That is, a function may be continuous but not differentiable.
Counterexamples:
The following functions are continuous at points specified against them but they are not differentiable at those points.
 f (z) = at any point
 g(z) = z^{2} at zero
 h(z) = Re z at any point.
Example 2.26
Find the regions of analyticity of the functions
 f(z) = e^{z}
 f(z) = (x – y)^{2} + 2i(x + y)
 f(z) = .
Solution
 f(z) = e^{z} = e^{x+iy} = e^{x}(cos y+sin y) = u+iv (say)
⇒ u = e^{x} cos y,
v = e^{x} sin y
⇒ u_{x} = e^{x} cos y,
u_{y} = −e^{x} sin y
v_{x} = e^{x} sin y,
v_{y} = e^{x} cos y
⇒ u_{x} = v_{y};
u_{y} = −v_{x}
u_{x}, u_{y}, v_{x}, v_{y} are continuous. CREs are satisfied for all x and y.
Hence e^{z} is analytic throughout the complex plane and it is an entire function.

u = (x − y)^{2};
v = 2(x + y)
u_{x} = 2(x − y);
u_{y} = −2(x − y);
v_{x} = 2, v_{y} = 2
u_{x} = v_{y} ⇒ x − y = 1
also u_{y} = −v_{x} ⇒ x – y = 1
CREs are satisfied only along the straight line x – y = 1.
f ′(z) = u_{x} + iv_{x} = 2(x – y) + 2i = 2 + 2i exists along x – y = 1 and not through any region (nbd) R. Hence f(z) is nowhere analytic.
 f(z) = is the quotient of polynomials z – 2i and z^{2} + 1 which are analytic throughout ℂ. So, f(z) is analytic everywhere except at points where z^{2} + 1 = 0, i.e., z = ±i.
Example 2.27
Show that z^{2} is analytic for all z.
Solution
f(z) = u + iv = z^{2} = (x + iy)^{2} = (x^{2} – y^{2}) + i (2xy)
⇒ u = x^{2} – y^{2}, v = 2xy
⇒ u_{x} = 2x; v_{y} = 2x; u_{y} = –2y; v_{x} = 2y
⇒ u_{x} = V_{y}, u_{y} = –V_{x}
u_{x}, u_{y}, v_{x}, v_{y} are continuous and CREs are satisfied; f(z) is analytic in ℂ.
Example 2.28
Show that f(z) = z + 2 is not analytic anywhere in the complex plane.
[JNTU 2000, 2003(2)]
Solution Let z = x + iy and
f(z) = u + iv = z + 2 = (x + iy) + 2(x – iy)
⇒ u = 3x, v = –y
⇒ u_{x} = 3, V_{x} = 0, u_{y} = 0, V_{y} = 1
CREs u_{x} ≠ v_{y}, u_{y} = –v_{u} are not satisfied.
Hence f(z) is not analytic anywhere.
Example 2.29
Show that = 0 where f(z) is an analytic function.
[JNTU 2002]
Solution Since = – , we have
Since f(z) is analytic f^{′}(z) is analytic. f^{′}() is also analytic
Example 2.30
If f(z) = u + iv is an analytic function of z prove that
[JNTU 2003S (4)]
Solution We know that
and
Differentiate first w.r.t taking f(z) as constant and then w.r.t z taking f() as constant; we have
Example 2.31
Find all values of k such that f(z) = e^{x}(cos ky + i sin ky) is analytic.
[JNTU 2003(4)]
Solution
Let f (z) = u + iv = e^{x} (cos ky + i sin ky)
⇒ u = e^{x} cos ky, 
v = e^{x} sin ky 
⇒ u_{x} = e^{x} cos ky, 
u_{y} = −ke^{x} sin ky 
v_{x} = e^{x} sin ky, 
v_{y} = ke^{x} cos ky 
CREs. u_{x} = v_{y} ⇒ k = 1, 
u_{y} = −v_{x} ⇒ k = 1 
f (z) is analytic ⇔ k = 1
Example 2.32
Prove that
where w = f(z) is analytic.
[JNTU 2004(4), 2004S, 2005]
Solution
Let f(z) = u + iv ⇒ Re f(z) = u.
Also f^{′} = u_{x} + iv_{x} so that
Squaring
Now
Similarly
We know that Re f(z) = u satisfies the Laplace’s equation so that
Example 2.33
Determine whether the function 2xy + i(x^{2} – y^{2}) is analytic.
[JNTU 2003(4)]
Solution
Let f (z) = u + iv = 2xy + i(x^{2} − y^{2})
u = 2xy; v = x^{2} − y^{2}
CREs are u_{x} = v_{y} and u_{y} = – v_{x} which are clearly not satisfied. Hence the given function is not analytic.
Example 2.34
Show that f(z) = xy^{2}(x + iy)/(x^{2} + y^{4}) (z ≠ 0) and f(0) = 0 is not analytic at z = 0 although CREs are satisfied at the origin.
Solution
Clearly,
Along the path y = mx,
Also, along the path x = my^{2},
Since the limit value depends on m and is different for different paths, the limit does not exist. Hence f(z) is not differentiable at z = 0. So, f(z) is not analytic at z = 0.
We now prove that CREs are not satisfied at z = 0.
Let
Also u(0, 0) = 0 and v(0, 0) = 0
∵ f (z) = 0 at z = 0
The CREs are satisfied at the origin z = 0. Consequently, f(z) is not analytic at (0, 0) though CREs are satisfied at z = 0.
Example 2.35
Show that the function f(z) = is not analytic at the origin, although CauchyRiemann equations are satisfied at the origin.
Solution Let f(z) = u(x, y) + iv(x, y) = so that u(x,y) = and v(x,y) = 0. We compute the partial derivatives at the origin.
Hence, Cauchy–Riemann equations , are satisfied at the origin.
We now find the derivative at the origin.
The limit must exist along any path leading to the origin. If we allow z = (x, y) → (0, 0) along the straight line y = mx, we obtain
which depends on m and hence not unique. Hence f^{′}(0) does not exist.
Note:
In this problem, all exist at (x, y) = (0, 0) and also satisfy Cauchy–Riemann equations at the origin. Yet f′(0) does not exist because the partial derivatives are not continuous at the origin.
Example 2.36
Prove that the function f (z) defined by
is continuous and the CauchyRiemann equations are satisfied at the origin but f^{′}(0) does not exist.
[VTU 2001, JNTU 2000, Punjab 1999]
Solution
Also, f(0) = 0 by hypothesis so that f(z) = f(0), when x → 0, y → 0 and y → 0, x → 0.
Now, let x and y both tend to 0 simultaneously along the path y = mx
Thus, irrespective the manner in which z → 0 we have f(z) = 0 = f(0), which implies that the function is continuous at z = 0.
Also,

u(0, 0) 



+ f (0) = 0. 

v(0, 0) 

Now, we complete partial derivatives for u, v at the
Clearly, Cauchy–Riemann equations , are satisfied at the origin.
Also,
f′(0) depends on m and is not unique. Therefore, f′(z) does not exist at the origin.
f(z) is not analytic at the origin even though it is continuous and satisfies Cauchy–Riemann equations at z = 0 because the partial derivatives are not continuous at the origin.
2.2.10 Orthogonal Trajectories
The curves u(x, y) = c_{1} and v(x, y) = c_{2} where f = u + iv ∈ H(D) represents two families of curves in the xyplane which are mutually orthogonal.
Suppose a member of
cuts a member of
at a point P(x_{0}, y_{0}).
By implicit differentiation of u(x, y) = c_{1} we get = slope of curves Eq. (1)
at P =
Similarly, differentiation of v(x, y) = c_{2} gives m_{2} = slope of curves (2) at P =
The product of the slopes of the curves at the intersection point P m_{1}. m_{2} =
.^{.}. u_{x} = v_{y}, u_{y} = –v_{x} by CREs.
Hence the family of curves u(x, y) = c_{1} and the family v(x, y) = c_{2} cut each other orthogonally.
Example 2.37
Find the orthogonal trajectories of
Solution The curves given by v(x,y) = constant will be the required orthogonal trajectories if f(z) = u + iv is analytic.
Differentiating (1) partially w.r.t. x and y
∴ v_{y} = u_{x} = 3x^{2}y – y^{3}, by Cauchy Riemann Eqs.
Integrating w.r.t y we get v =
Differentiating w.r.t x, we get
where c_{1} is an arbitrary constant. The required orthogonal trajectory is
Example 2.38
Prove that u(x, y) = x^{3} – 3xy^{2} = c_{1} and v(x, y) = 3x^{2}y – y^{3} = c_{2} cut each other orthogonally.
[JNTU 2003(3)]
Solution Let a member of the family
intersect a member of the family
Differentiating Eqs.(1) and (2), we get
Clearly the product of their slopes at P = –1 and hence the problem.
Example 2.39
Find the orthogonal trajectories of the family of the curves r^{2} cos 2θ = c.
Solution Let u(r, θ) = r^{2} cos 2θ = c
on integrating v(r, θ) = r^{2} sin 2θ + c_{1}
Differentiating v w.r.t r, we get
Also = constant.
∴ The orthogonal trajectories of curves u(r, θ) = r^{2} cos 2θ = c are the curves v(r, θ) = r^{2} sin 2θ = c.
EXERCISE 2.2
 Verify CREs for the functions f(z)=
 z^{2},
 iz + ,
 sin z.
 Prove that the following functions are nowhere differentiable:
 f(z) = z,
 f(z) = Re ,
 f(z) = 2x + ixy^{2}.
 Prove that the following functions are differentiable at every point:
 f(z) = x^{2} – y^{2} – 2xy + i(x^{2} – y^{2} + 2xy)
 f(z) = 2x – 3y + i(3x + 2y)
 f(z) = iz + 2.
 Prove that f(z) = cosx(cosh y + a sinh y)+ i sin x (cosh y + b sinh y) is analytic if a = b = –1.
 Determine constants a, b and c such that the following functions are differentiable at every point:
 f(z) = x + ay + i(bx + cy)
 f(z) = e^{x} cos ay + i. e^{x} sin(y + b) + c.
 Prove that the functions f(z) and f() are simultaneously analytic.
2.3 Laplace’s Equation: Harmonic and Conjugate Harmonic Functions
The real part u(x, y) and the imaginary part v(x, y) of an analytic function f(z) = u(x, y) + iv(x, y) satisfy the most important differential equation in physics:
Laplace’s equation
(∇ is called nabla or del) (viz.) ∇^{2}u = u_{xx} + u_{yy} = 0 in D ∇^{2}v = v_{xx} + v_{yy} = 0 in D.
Equation ∇^{2}F = 0 occurs in gravitation, electrostatics, fluid flow, heat conduction and so on.
Theorem 2.5 (Laplace’s equation) If f(z) = u(x, y) + iv(x, y) is analytic in a domain D then u and v satisfy Laplace’s equation
in D and have continuous second partial derivatives in D.
Proof: Differentiating u_{x} = v_{y} w.r.t x and u_{y} = – v_{x} w.r.t y, we have
The derivative of an analytic function is itself analytic ⇒ u, v have continuous partial derivatives of all orders.
In particular, v_{yx} = v_{xy}.
Adding Eq. (2.20) we obtain Eq. (2.18). Equation (2.19) is similarly proved.
Solutions of Laplace’s equation having continuous secondorder partial derivatives are called harmonic functions and their theory is called Potential Theory.
Hence the real and imaginary parts of an analytic function are harmonic functions.
If two harmonic functions u and v satisfy CREs in a domain D they are the real and imaginary parts of an analytic function f in D. v is said to be a conjugate harmonic function of u in D and vice versa.
Let f(z) = u + iv be an analytic function whose real part u is known. We can find the imaginary part v and also the function f(z) by following either of the two methods:
Method 1: Exact Differential Method using CREs u is a harmonic function
Let
where by Eq. (2.22)
Now



= −∇^{2}u = 0 by Eq. (2.21) 
is an exact differential equation.
Hence integrating Eq. (2.23) we can find v. Having known u and v, f(z) = u + iv can be determined.
Method 2: Milne–Thomson’s method
Since z = x + iy, = x − iy we have
Considering Eq. (2.24) as a formal identity in z and and putting = z, we get
which is obtained from f (z) = u(x, y) + iv(x, y) by replacing x by z and y by 0.
Now
If we set and then,
Replacing x by z and y by 0 in Eq. (2.26), we obtain
f ′(z) = Ф_{1} (z, 0) – iФ_{2}(z, 0)
⇒f (z) = [Ф_{1} (z, 0) – iФ_{2}(z, 0)] + c_{1}
where c_{1} is a complex constant.
Similarly if v(x, y) is given, we can find u such that u + iv is analytic.
where
replacing x by z and y by 0.
Integrating we get
where c_{2} is a complex constant.
Example 2.40
Verify that u = x^{2} – y^{2} – y is harmonic in C and find a conjugate harmonic function v of u.
Solution ∇^{2}u = 0 by direct calculation. Now u_{x} = 2x, u_{y} = –2y – 1. Hence, because of CREs a conjugate v of u must satisfy the following:
v_{y} = u_{x} = 2x; (1)
v_{x} = −uy = 2y + 1 (2)
Integrating Eq. (1) w.r.t y and differentiating the result w.r.t x, we get
Comparison with Eq. (2) shows that dh / dx = 1 ⇒ h(x) = x + c. Hence v = 2xy + x + c (c any real constant) is the most general conjugate harmonic of the given function u. The corresponding analytic function is
f (z) 
= 
u + iv = x^{2} − y^{2} − y + i(2xy + x + c) 

= 
z^{2} + iz + ic 
The conjugate of a harmonic function is uniquely determined up to an arbitrary real additive constant.
Harmonic Functions
Example 2.41
If u(x, y) and v(x, y) are harmonic functions in a region R, prove that the function is an analytic function.
[JNTU 2003S(1)]
Solution u and v are harmonic functions
Let
Differentiating Eqs. (3) and (4) partially w.r.t x and again w.r.t y, we get
Now by Eq. (2)
by Eq. (1)
Thus U, V satisfy CREs and their partial derivatives U_{x}, U_{y}, V_{x}, V_{y} are continuous.
Hence is an analytic function.
2.3.1 Harmonic and Conjugate Harmonic Functions
Example 2.42
Show that u = 2 log(x^{2} + y^{2}) is harmonic and find its harmonic conjugate.
[JNTU 2006(1)]
Solution u = 2 log(x^{2} + y^{2})
Adding we get
⇒ u is an harmonic function.
If v(x, y) is the harmonic conjugate of u(x, y) then
Integrating we get
Example 2.43
Show that if u and v are conjugate harmonic functions then the product uv is a harmonic function.
Solution u, v are conjugate harmonic functions
⇒ u_{xx} + u_{yy} = 0, v_{xx} + v_{yy} = 0 (1), (2)
u_{x} = v_{y}, u_{y} = − v_{x} (3), (4)
If Φ 
= 
uv then Φ_{x} = uv_{x} + vu_{x}, 
Φ_{xx} 
= 
uv_{xx} + 2u_{x}v_{x} + vu_{xx} 
and Φ_{yy} 
= 
uv_{yy} + 2u_{y}v_{y} + vu_{yy} 

= 
uv_{yy} − 2v_{x}u_{x} + vu_{yy}, by CREs 
Now Ф_{xx} + Ф_{yy} = u(v_{xx} + v_{yy}) + v(u_{xx} + u_{yy}) = 0.
Hence Ф = uv is a harmonic function.
Example 2.44
Show that v(x, y) = – sin x sinh y is harmonic. Find the harmonic conjugate of v.
Solution Differentiating v partially w.r.t x and y, we get
v_{x} = −cos x sinh y, 
v_{xx} = sin x sinh y 
v_{y} = −sin x cosh y, 
v_{yy} = −sin x sinh y 
∇^{2}v = v_{xx} + v_{yy} = 0 
∴ v is harmonic. 
To find u (harmonic conjugate of v), by CREs,
Integrating Eq. (1) partially w.r.t x, we get
where h(y) is an arbitrary function of y
Differentiating Eq. (3) partially w.r.t y we get
u_{y} = cos x sinh y + h′(y) = −v_{x} = cos x sinh y
⇒ h′(y) = 0 ⇒ h(y) = constant
∴ u(x, y) = cos x cosh y + c
The required analytic function is
f (z) 
= 
cos x cosh y + c + i(− sin x sinh y). 

= 
cos(x + iy) + c = cos z + c. 
Example 2.45
Prove that u = 2x – x^{3} + 3xy^{2} is harmonic and find its harmonic conjugate. Find also the corresponding analytic function.
Solution Let u = 2x – x^{3} + 3xy^{2}
⇒ u_{x} = 2 − 3x^{2} + 3y^{2}, u_{xx} = −6x,
uy = 6xy, u_{yy} = 6x
⇒ u is harmonic
Let v be a harmonic conjugate of u so that f (z) = u + iv is analytic.
By CREs we have v_{y} = u_{x} = 2 – 3x^{2} + +3y^{2}
Integrating w.r.t y we get
where A(x) is an arbitrary function of x.
Differentiating Eq. (1) partially w.r.t x, we get v_{x} = –6xy + A′(x) = –u_{y} = –6xy ⇒ A′(x) = 0 so that A(x) = c a pure complex constant
⇒ v = 2y − 3x^{2}y + y^{3} + c
Now f(z) 
= 
(2x − x^{3} + 3xy^{2}) + i(2y − 3x^{2}y + y^{3}) + ic 

= 
2(x + iy) − [(x^{3} − 3xy^{2}) + i(3x^{2}y − y^{3})] + ic 

= 
2z − z^{3} + ic. 
Example 2.46
Show that v(x, y) = – sin x sinh y is harmonic. Find the harmonic conjugate of v.
Solution Differentiating v(x,y) partially w.r.t x and y, we get
∴ v is a harmonic function in D
To find u, the harmonic conjugate of v:
By CREs u_{x} = v_{y} = − sin x cosh y, 3(a)
u_{y} = − v_{x} = cos x sinh y 3(b)
Integrating 3(a) partially w.r.t x, we get
where h(y) is an arbitrary function of y.
Now differentiating Eq. (4) w.r.t y we get
uy 
= 
cos x sinh y + h′(y) 

= 
−v_{x} = cos x sinh y by Eq. 3(b) 
⇒ h′(y) 
= 
0 ⇒ h(y) = c (const.) 
The required analytic function is
f (z) 
= 
u(x,y) + iv(x,y) 

= 
cos x cosh y − i sin x sinh y + c 

= 
cosz + c 
MilneThomson’s method
Example 2.47
Find the analytic function f (z) = u + iv where u(x, y) = e^{x}(x cos y – y sin y) + 2 sin x sinh y + x^{3} – 3xy^{2} + y.
Solution Differentiating u(x, y) partially w.r.t x and y, we get
u_{x} 
= 
e^{x} (x cos y − y siny + cosy) + 2 cos x sinh y + 3x^{2} − 3y^{2} 
u_{y} 
= 
e^{x} (−x sin y − siny − y cosy) + 2 sinx coshy − 6xy + 1 
We know that f′(z) = u_{x} + iv_{x} = u_{x} – iu_{y},
By CREs v_{x} = – u_{y}
Replace x by z and y by 0 to obtain f′(z)
∴ f ′(z) 
= 
e^{z} (z · 1 − 0 + 1) + 0 + 3z^{2} − 0 − i[e^{z} (0 − 0 − 0) + 2 sinz · 1 − 6z 0 + 1] 

= 
e^{z} (z + 1) + 3z^{2} − 2i sin z − i 
Integrating, we obtain
Example 2.48
Show that u(x, y) = sin x cosh y + 2 cos x sinh y + x^{2} – y^{2} + 4xy is harmonic. Find the analytic function f (z) with u as its real part.
Solution Differentiating u partially w.r.t x and y twice, we get
u_{x} 
= 
cos x cosh y − 2 sin x sinh y + 2x + 4y 
u_{xx} 
= 
− sin x cosh y − 2 cos x sinh y + 2 
u_{y} 
= 
sin x sinh y + 2 cos x cosh y − 2y + 4x 
u_{yy} 
= 
sin x cosh y + 2 cos x sinh y − 2 


Hence u is harmonic.
Let Φ_{1}(x,y) 
= 
u_{x} and Φ_{2} (x,y) = u_{y} 
so that Φ_{1} (z, 0) 
= 
cosz cosh 0 − 2 sinz sinh 0 + 2z 

= 
cos z + 2z 
and Φ_{2} (z, 0) 
= 
sin z sinh 0 + 2 cos z cosh 0 − 2 × 0 + 4z 

= 
2 cos z + 4z 
by Milne–Thomson’s method, where c is an arbitrary complex constant
Example 2.49
If u(x,y) = sin 2x/(cosh 2y + cos 2x) is the real part of analytic function f (z) = u(x, y) + iv(x, y). Find f (z).
Solution We can easily verify that u_{xx} + u_{yy} = 0 and hence u is harmonic. Differentiating u partially w.r.t x and y we get
Example 2.50
Find the analytic function f (z) = u + iv if
Solution
Since f (z) = u + iv is an analytic function
Replacing x by z and y by 0,
Eq. (3) ⇒
Replacing x by z and y by 0,
Adding Eqs. (4) and (5), we get
Subtracting Eqs. (4) from (5), we get
Now f (z) 
= 
u(z, 0) + iv(z, 0) 
⇒ f ′(z) 
= 
u_{x} (z, 0) + iv_{x} (z, 0) 
⇒ f ′(z) 
= 
u_{x} (z, 0) − iv_{y} (z, 0) 


Integrating w.r.t z, we obtain
where c is an arbitrary complex constant.
Example 2.51
If u + v = (x – y)(x^{2} + 4xy + y^{2}) find the analytic function f (z) = u + iv.
Solution
f(z) 
= 
u + iv 
⇒ (1 + i)f (z) 
= 
(u − v) + i(u + v) 

= 
U + iV (say) 
Now V 
= 
u + v = (x − y)(x^{2} + 4xy + y^{2}) 
By MilneThomson’s method,
where c is an arbitrary complex constant.
Example 2.52
If v(x,y) = x^{4} – 6x^{2}y^{2} + y^{4} find f (z) = u(x, y) + iv(x, y such that f (z) is analytic.
Solution Differentiating v partially w.r.t x and y twice, we get

v_{x} = 4x^{3} − 12xy^{2} = −u_{y} 
(1) 

v_{y} = −12x^{2}y + 4y^{3} = u_{x} 
(2) 

v_{xx} = 12x^{2} − 12y^{2}) 
(3) 

v_{yy} = −12x^{2} + 12y^{2} 
(4) 
⇒ ∇^{2}v = 0 ⇒ v is harmonic
Integrating Eq. (2) w.r.t x, we get
where A(y) is an arbitrary function of y Differentiating w.r.t to y we get
u_{y} 
= 
−4x^{3} + 12xy^{2} + A′(y) = −v_{x} 

= 
−4x^{3} + 12xy^{2} ⇒ A′(y) = 0 
A(y) 
= 
c = constant 
∴ u(x,y) 
= 
−4x^{3}y + 4xy^{3} + c 
Hence f(z) 
= 
(−4x^{3}y + 4xy^{3} + c) + i (x^{4} − 6x^{2}y^{2} + y^{4}) 

= 
i[(x^{4} − 6x^{2}y^{2} + y^{4}) + i(4x^{3}y − 4xy^{3})] + c 

= 
i(x + iy)^{4} + c = iz^{4} + c 
MilneThomson’s method:
Let Ψ_{1} (x,y) 
= 
v_{y} = −12x^{2}y + 4y^{3} 
and Ψ_{2}x,y 
= 
v_{x} = 4x^{3} − 12xy^{2} 
Ψ_{1} (z, 0) 
= 
0 and Ψ_{2} (z,0) = 4z^{3} 
where c is an arbitrary complex constant.
Example 2.53
Show that u = sin x cosh y + 2 cos x sinh y + x^{2} – y^{2} + 4xy satisfies the Laplace equation. Find the corresponding analytic function.
[JNTU 2008]
Solution It is given that
u(x,y) = sin xcosh y + 2 cos x sinh y + x^{2} − y^{2} + 4xy (1)
Differentiating Eq. (1) twice partially w.r.t x and y, respectively, we get
u_{x} = cos x cosh y − 2 sin x sinh y + 2x + 4y 
(2) 
u_{xx} = − sin x cosh y − 2 cos x sinh y + 2 
(3) 
u_{y} = sin x sinh y + 2 cos x cosh y − 2y + 4x 
(4) 
u_{yy} = sin x cosh y + 2 cos x sinh y − 2 
(5) 
Adding Eqs. (3) and (5), we obtain
u satisfies Laplace’s equation and hence u is a harmonic function.
Also,
Replace x by z and y by 0 (MilneThomson’s method)
⇒ cos x cosh y − 2 sin x sinh y + 2x + 4y − i(sin x sinh y + 2 cos x cosh y − 2y + 4x)
= Φ_{1}(z, 0) − iΦ_{2}(z, 0) = Φ(z) (say)
= (cos z · 1 − 0 + 2z + 0) − i(sin z · 0 + 2 cos z · 1 − 2 × 0 + 4z)
Or Φ(z) = (1 − 2i) cos z + 2(1 − 2i)z
Integrating w.r.t z,
Example 2.54
Find the analytic function whose imaginary part is e^{–x} (x cos y + y sin y).
[JNTU 1998S]
Solution Let f(z) = u + iv be the analytic function. Then v = e^{–x} (x cos y + y sin y)
We know that f′(z) 
= 
u_{x} + iv_{x} = v_{y} + iv_{x} 

= 
Φ_{1} + iΦ_{2} = Φ(z) say 
Since f (z) 
= 
u + iv ∈ H(D), 
u_{x} 
= 
u_{y}, u_{y}, = −v_{x} (CREs) 
Now v_{x} 
= 
e^{−x} (−x cos y − y sin y + cos, y 

= 
Φ_{2}(z) = e^{−z} (−z + 1) 
v_{y} 
= 
e^{−x}(−x sin y + sin y + y cos y) 

= 
Φ_{1}(z) = e^{−z} (0) 
Φ(z) 
= 
Φ_{1}(z,0) + iΦ_{2}(z,0) = i(i −z)e^{−z} 
Integrating w.r.t z, we get
Example 2.55
Find the analytic function whose real part is u = e^{x} [(x^{2} – y^{2}) cos y – 2xy sin y].
[JNTU 2008]
Solution
u 
= 
e^{x}[(x^{2 }− y^{2}) cos y − 2xy sin y] 
⇒ u_{x} 
= 
e^{x}[(x^{2} − y^{2}) cos y − 2xy sin y + 2x cos y − 2y sin y] = Φ_{1} (x, y) 
u_{y} 
= 
e^{x}[−2y cos y − (x^{2} − y^{2}) sin y − 2x sin y −2xy cos y] = Φ_{2}(x, y) 
∴ f′ (z) 
= 
u_{x} + iv_{x} = u_{x} − iu_{y} 

= 
Φ_{1}(z, 0) − iΦ_{2}(z, 0) 

= 
e^{z} (z^{2} + 2z) − e^{z} (0) = e^{z} (z^{2} + 2z) 
Integrating, f (z) = z^{2}e^{z} + c
Example 2.56
If f (z) = u + iv ∈ H (D) then prove that
 ∇^{2}f (z)^{2} = 4f′(z)^{2}.
JNTU 1993]
 ∇^{2}u^{p} = p(p − 1)u^{p−2}f′(z)^{2}.
[JNTU 1997S]
Solution Since f (z) = v + iv is analytic in a domain D, the functions u = u(x, y) and v = v(x,y) satisfy CauchyRiemann equations in D
Further, u and v are harmonic functions in D which implies that u, v satisfy Laplace’s equation ∇^{2} Φ = 0, i.e., ∇^{2}u = 0, ∇^{2}u = 0 in D
 f (z) = u + iv ⇒ f (z)^{2} = u^{2} + v^{2}
We calculate partial derivatives w.r.t x and y of f (z)^{2} = u^{2} + v^{2}
Similarly,
Adding the two results (3) and (4)
by Eq. (2)by Eq. (1)= 4u_{x} − iu_{y}^{2}
= 4f′(z)^{2}
(5)
 Differentiating u^{p} partially w.r.t x twice, we get
Similarly, we obtain
Adding Eqs. (6) and (7), we obtain
EXERCISE 2.3
 Prove that the following functions are harmonic. Find their harmonic conjugates.
 u = x^{2} – y^{2} – y,
Ans: v = 2xy + x + c.

Ans:
 u = 3xy^{2} – x^{3},
Ans: v = y^{3} – 3x^{2}y + c
 v = y^{2} – x^{2},
Ans: u = 2xy + c
 u = x^{2} – y^{2} – y,
 Find the analytic function f (z) =u + iv in question 1.
Ans:
 z^{2} + iz + c
 −z^{3} + ic
 −iz^{2} + c
 Find the analytic function f (z) = v + iv such that and f (π / 2) = (3 − i) / 2.
Ans: f (z) = cot(z/2) + (1 − i)/2
 Find the analytic function f (z) = u + iv such that and f (π / 2) = 0.
Ans:
 Find the analytic function f (z) = u + iv such that
Ans:
 If f (z) = u + iv is an analytic function of z = x − iy and u − v = c^{x} (cos y − sin y) find f (z) in terms of z.
[JNTU 1995S]
Ans: f (z) = e^{z} + c(1 + i)
 Show that u(x, y)= sin x cosh y + 2 cos x sinh y + x^{2} − y^{2} + 4xy is harmonic. Find an analytic function f (z) with u as its real point.
Ans: f (z) = sin z + z^{2} − 2i sin z − 2iz^{2} + c
 Find the real part of the analytic function whose imaginary part is e^{x}[2xy cos y + (y^{2} − x^{2}) sin y]. Find the analytic function.
Ans: u = e^{−x}[(x^{2} − y^{2}) cos y + 2xy sin y]
f (z) = e^{−z} · z^{2}
 Prove that u + iv is analytic if u = Φ_{y} − ψ_{x} and v = Φ_{x} − ψ_{y} where Φ and ψ are harmonic function.
 If f (z) is analytic, prove that
 If f(z) = u + iv is analytic prove that and ∇^{2}(amp f (z)) = 0.
 From Laplace’s equation for u(x, y) prove that .
 Show that f (x, y) = x^{3}y − xy^{3} + xy + x + y can be the imaginary part of an analytic function of z = x + iy.
 Find the analytic function f (z) = u(r, θ) + iv(r, θ) such that
 u(r, θ) = −r^{3} sin 3θ
 .
Ans:
 iz^{3} + ic
 If show that the curves u(x, y) = c_{1} and v(x, y) = c_{2} intersect orthogonally.