Question Papers
Code No.R059210201 
Set No.1 
Time: 3 hours 
Max Marks: 80 
1.
 Evaluate dx using β − γ functions
 Evaluate using β − γ functions
 Evaluate dx using B, Г functions
[5+6+5]
Solution
 Put log
The limits for t are ∞, 0.
 Put
The limits for t are 0, ∞.
Note:
The upper limit should be ‘a’ and not −1 as printed.

Put x = a sinθ, dx = a cosθ dθ
The limits for θ are 0, π/2.
2.
 Show that
Solution

Putting n = 1, 2, 3, we get from Eq. (1)
Substituting for J_{2}(x) from Eq. (2) into Eq. (3), we get
Substituting for J_{2}(x) from Eq. (5) and J_{2}(x) from Eq. (2) into Eq. (4) we get
 P_{n}(x) is a solution of Legendre’s equation
(1 − x^{2})y″ − 2xy′ + n(n + 1)y = 0 or
[(1 − x^{2})y′]′ + n(n + 1)y = 0 (1)
⇒ n(n + 1)P_{n}(x) = −[(1 − x^{2})P′_{n} (x)]′ (2)
Multiplying both sides by P_{n}(x) and integrating by parts w.r.t x from − 1 to 1
3.
 If w = u + iv is an analytic function of z and
 If sin(θ + iα) = cos α + i sin α, then prove that cos^{4} θ = sin^{2} α.
[8+8]
Solution
 Please refer to Ex. 2.45 in Ch. 2.
 sin(θ + iα) = sin θ cosh α + i cosθ sinh α
= cos α + i sin α (1)
Equating the real and imaginary parts, we have
Squaring and adding, we get
⇒ cos^{2} α − cos^{2} θ − sin^{2} α sin^{2} θ = sin^{2} θ cos^{2} θ
⇒ (1 − sin^{2} α) cos^{2} θ − sin^{2} α(1 − cos^{2} θ)
= (1 − cos^{2} θ) cos^{2} θ ⇒ cos^{4} θ = sin^{2} α.
4.
 Find f(2) and f(3) if f(a) = where C is the circle z = 2.5 using Cauchy’s integral formula
 Evaluate where C is the circle z = 1 using Cauchy’s integral formula.
Solution
 a = 2 lies inside C: z = 2.5, circle of radius 2.5. By Cauchy’s integral formula,
a = 2 lies inside C: z = 2.5, circle of radius 2.5. By Cauchy’s integral formula,
since the integrand is analytic inside C.

Put x = e^{iθ} log z = log e^{iθ} = iθ; dz = ie^{iθ}dθ
The limits for θ are 0 to 2π
5.
 Find the Laurent expansion of for 1 < z < 3
 Expand in a Laurent series for z < 3.
[8+8]
Solution
 Putting into partial fraction, we have for 1 < z < 3 the Laurent series expansion is

6.
 Find the poles and the residues at the poles of
 Evaluate where C is z = 5 by residue theorem.
[8+8]
Solution
 The poles of f(z) = are the zeros of the denominator z^{2} + 1 = 0
⇒ x = i, −i, which are simple poles of f(z)
 The poles of f(z) = are the zeros of the denominator z_{2} + 9 = 0, i.e., z = 3i, −3i which are simple poles of f(z) lying inside z = 5
By Cauchy’s residue theorm,
7.
 Show that (a^{2} < 1) using Residue Theorem
 Show by the method of contour integration that
Solution
 Please refer Ex. 6.28 in Ch. 6.
 Let C be the closed contour consisting of
 Line Segment L [R,R] along the real axis
 Semicircle S_{R} : z = Re^{iθ} (0 ≤ θ ≤ π) in the upper half plane and lef f(z) = Then f(z) has double poles at z = ±ai of which z = ai alone is inside C. By Cauchy’s Residue Theorem
Therefore from (1) – (3) R → ∞, we obtain
Since the integrand is an even function of x in (−∞,∞)
8.
 Show that image of the hyperbola x^{2} − y^{2} = 1 under the transformation w = 1/z is r^{2} = cos 2θ
 Show that the transformation changes the circle x^{2} + y^{2} − 4x = 0.
[8+8]
Solution
 Let z = re^{iθ}, w = Re^{iϕ}. The transformation becomes
The hyperbola x^{2} − y^{2} = 1 becomes
r^{2} cos^{2} θ − r^{2} sin^{2} θ = 1
⇒ r^{2}(cos^{2} θ − sin^{2} θ) = 1 or r^{2} cos2θ = 1
⇒ cos(−2ϕ) = 1 by Eq. (1)
⇒ R^{2} = cos2ϕ
Thus the hyperbola x^{2} − y^{2} = 1 under the transformation becomes lemniscate R^{2} = cos 2ϕ.
 Solving for z, we have
Now the equation of the circle x^{2} + y^{2} − 4x = 0 is
Thus the image of the circle is obtained by substituting Eqs. (1) and (2) into Eq. (3)
Code No.R059210201 
Set No.2 
Time: 3 hours 
Max Marks: 80 
1.
 Show that β(m, n) = γ(m)γ(n)/γ(m + n)
 Show that where n is an odd integer
 Show that
Solution
 Please refer Theorem 6.1 in Ch. 6.
 Put x = sin θ, dx = cos θdθ, The limits for θ are 0 and π/2.
since n is odd put n = 2m + 1
2.
 Establish the formula P′_{n+1}(x) − P′_{n−1}(x) = (2n + 1)P_{n}(x)
 Prove that
 When n is an integer, show that J_{−n}(x) = (−1)^{n}J_{n}(x).
[6+5+5]
Solution
 Please refer to RR1 (Legendre polynomisdl) Ch. 1.
 Please refer to RR2 (Bessel functions) Ch. 1.
 Please refer to the proof of Ex. 1.57(i) in Ch. 1 (Bessel functions.)
3.
 Find the analytic function whose imaginary part is f(x, y) = x^{3}y − xy^{3} + xy + x + y where z = x + iy
 Prove that where w = f(z) is analytic.
[8+8]
Solution
 Let
ϕ(z) = g(x, y) + if (x, y) ∈ H(D) (1)
whose imaginary part is
f(x, y) = x^{3}y − xy^{3} + xy + x + y (2)f is harmonic function in D ⇒ f_{xx} + f_{yy} = 0. we have,
f_{x} = 3x^{2}y − y^{3} + y + 1 and
f_{y} = x^{3} − 3xy^{2} + x + 1
f_{xx} = 6xy and f_{yy} = − 6xy
⇒ f_{xx} + f_{yy} = 0 (3)
Also, f, g satisfy CREs:
g_{x} = f_{y} and g_{y} = − f_{x} (4)by MilneThomson’s rule of replacing x by z and y by 0.
Integrating Eq. (8) w.r.t z, we get
 Let f(z) = u + iv ∈ H(D)
Then Re f(z) = u (1)
Also
Similarly
Adding Eqs. (3) and (4), we get
[by Eq. (2)] (5)
4. (a),(b) Please refer to Question 4(a), (b) and its Solution in JNTURegular Exam Nov. 2008 (Set 1).
5.
 State and prove Laurent’s theorem.
 Obtain all the Laurent series of the function about z = −2.
[8+8]
Solution
6.
 Find the poles and residue at each pole of
 Find the poles and residue at each pole of
 Evaluate where C is z = 3/2.
Solution
 The poles of are the zeros of the denominator
 The poles of are the zeros of the denominator (z − 1)^{3} = 0 ⇒ z = 1 is a pole of order 3. The formula for finding residue is
 The poles of are the zeros of (z − 1)(z − 2) = 0, i.e., z = 1 lies inside C, circle of radius 3/2.
7.
 Show that (a ≥ b ≥0) using residue theorem.
 Evaluate, by contour integration,
[8+8]
Solution
 We evaluate by converting it into a contour integral taken over the unit circle z = 1 by putting and
where
Then
The poles are simple and at α and β (α ≤ β) of which a lies inside C while lies outside C.
By Cauchy’s residue theorem,
 Consider where C consists of the line segment [R, R] along the real axis and the semicircle C_{R} : z = R in the upper halfplane. has simple poles at z = ±i of which z = i lies inside C and z = −i outside.
By Cauchy’s residue theorem,
since z = x, dz = dx along the real axis
Also, on C_{R}: z = Re^{iθ}, bz = C_{R}: z = Re^{iθ} dθ and limits for θ are 0, π. Therefore we have,
8.
 Show that the transformation maps the interior of the circle z = 1 into the upperhalf of the wplane, the upper semicircle into the positive half of the real axis and the lower semicircle into the negative half of the real axis.
 By the transformation w = z^{2}, show that the circle z − a = c (a and c are real) in the zplane corresponds to the binacon in the wplane.
Solution
 The given transformation is
This can be written as By Componendodividondo this gives Writing w = u + iv and taking modulus, we have
This show that the mapping transforms the interior of the unit circle z ≤  into the upper half wplane.
Also z = 1 (the unit circle) can be defined by z = e^{iθ} (0 ≤0 ≤ 2π)
The transformation (1) gives
on multiplying the numerator and denominator by 2e−e^{iθ/2}
Equating the real and imaginary parts, we obtain u = tan θ/2, v = 0
u ≥ 0, v = 0 for 0 ≤ θ, 2 ≤ π/2 or 0 ≤ θ ≤ π (upper semicircle in the zplane)
u ≤ 0, v = 0 for + π/2 ≤ θ, 2 ≤ π or π ≤ θ ≤ 2 π (lower semicircle in the zplane)
Thus, the transformation (1) maps the upper semicircle in the zplane into the positive half of the real axis (u ≥ 0, v = 0) and the lower semicircle in the zplane into the negative half of the real axis (u ≤ 0, v = 0).
 The equation of the circle z − a = c may be written as z − a = ce^{iθ} ⇒ z = a + ce^{iθ}
The mapping w = z^{2} becomes
By taking the pole (origin) in the wplane at w = a^{2} − c^{2}, we put w − (a^{2} − c^{2}) = Re^{iθ} we have
Re^{iϕ} = 2ce^{iθ}(a + c cos θ)Equating the modulus and amplitude of the complex quantity on either side
R = 2c(a + c cos θ) and ϕ = 0so that R = 2c(a + ccos ϕ). This equation represents a binacon in the wplane. Thus the circle z − a = c in the zplane corresponds to binacon in the wplane.
Code No.R059210201 
Set No.3 
Time: 3 hours 
Max Marks: 80 
1.
 Show that
 Prove that
 Show that
[5+5+6]
Solution
 By definition,
Put x = sin^{2} θ ⇒ dx = 2 sin θ cos θdθ. The limits for θ are 0 and π/2. Now, Eq. (1) becomes
Taking in turn, we obtain from (3)
But we know that
 We know that,
Taking q = 1 − p we get from (5) and (6)
From calculus of residues, we have the result
2.
 Prove that
[16]
Solution Please refer to the orthogonal property of Legendre polynomials in Ch. 1
3
 Find the analytic function whose imaginary part is f(x,y) = x^{3}y − xy^{3} + xy + x + y where z = x + iy
 Prove that where f(z) is analytic
[8+8]
Solution
 Let f(z) = u+iv where v = x^{3}y−xy^{3}+xy+x+y Differentiating v w.r.t x and y, we get
f′(z) = u_{x} + iv_{x} = v_{y} + iv_{x}
= (x^{3} − 3xy^{2} + x + 1) + i(3x^{2}y − y^{3} + y + 1)
(by CauchyRiemann Equations)
= (z^{3} − 0 + z + 1) + i(0 − 0 + 0 + 1),
replacing x by z and y by 0,
according to Milne Thomson’s method= z^{3} + z + 1 + i
Integrating w.rtz, f(z) = 1/4z^{4} + 1/2z^{2} + z + iz + c, where c is a complex constant, is the required function.
 We have
4.
 Evaluate with C : z = 2 using cauchy’s integral formula.
 Evaluate
 Evaluate where C : z = 2 using Cauchy’s integral theorem.
[5+5+6]
Solution
 f(z) = (z^{3} − sin 3z) is analytic inside the circle C : z = 2 and the singular point a = π/2 of the integrand lies inside C.
C is the circle z = 2. The singularity z = 1 lies inside C.f(z) = e^{−z} is analytic f″(z) = e^{−z} ⇒ f″(1) = e^{−1}.
5.
 Expand about z = 1 as a Laurent series. Also find the region of convergence.
 Find Taylor series for about z = 1; also find the region of convergence.
Solution
 Put z − 1 = w or z = w + 1. Then
which is the region of convergence.
 z = −2 is the singularity of If the centre of the circle is at z = 1, then the distance of the singularity z = −2 from the centre is 3 units. If a circle of radius 3 units is drawn with its centre at z = 1, then f(z) is analytic within the circle z −1 = 3 and there fore we can expand f(z) in a Taylor’s series.
The region is the interior of the circle z − 1 = 3 and there fore we can expand f(z) in a Talylor’s series. The region is the interior of the circle z − 1 = 3.
6.
 Find the poles and the residue at each pole of
 Evaluate where C is z = 5, by residue theorem.
[8+8]
Solution
 Please refer to the Solution of 6(a) in JNTU Feb 2008 (Set 1).
 Please refer to Solution of 6(b) in JNTU Feb 2008 (Set 1).
7.
 Evaluate using residue theorem.
 Evaluate using residue theorem.
Solution We convert the integral into a contour integral taken around the unit circle C : z = 1 by putting z = e^{iθ} (0 ≤ θ ≥ 2π),
Poles of the integrand are , −2 of which lies inside C
by considering the contour integral ∫_{C}f(z)dz where f(z) = and C is the closed path consisting of the semicircel C_{R} : z = Re^{iθ} (0 ≤ θ ≤ π) and the line segment L : [−R, R] along the real axis. The integrand f(z) = has simple poles at z = ±i and ±2i, by equating to zero the denominator (z^{2} + 1) (z^{2} + 4), of which z = i, 2i lie within C in the upper half of the plane.
By residue theorem,
Now on C_{R} : z = Re^{iθ} (0 ≤ θ ≤ π) dz = iRe^{iθ} dθ
Letting R → ∞ we obtain from Eqs. (5) and Eqs. 6
8.
 Find the image of the infinite strip 0 < y < under the transformation w = 1 /z
 Find the bilinear transformation which maps the points (−1, 0, 1) into the points (0, i, 3i).
[8+8]
Solution
 Let z = x + iy, w = u + iv so that the transformation becomes
Equating the real and imaginary parts, we obtain from Eq. (1)
which is a circle of radius unity with centre (0, − 1).
∴ Under the transformation the straight line y = 0 (real axis) is transformed into v = 0 (real axis). Also, the straight line is transformed into the circle u^{2} + (v + 1)^{2} = 1. Thus the infinite strip 0 < y < in the zplane is mapped into the region between the line v = 0 and the circle u^{2} + (v + 1)^{2} = 1 under the mapping w =
 Let the transformation be
Substituting in Eq. (1), we get
by componendodividendowhich is the required transformation.
Set No.4 
Time: 3 hours 
Max Marks: 80 
1.
 Evaluate using β − γ
 Prove that using β − γ function and evaluate.
 Show that
[5+5+6]
Solution
[by putting x^{2} = t ⇒ x = dt and the limits for t are 0, ∞]
[by putting x^{4} = t ⇒ x = dt and the limits for t are 0, ∞]
 Put x = at, dx = adt and the limits for t are 0, ∞.
2.
 Show that the coefficient of t^{n} in the power series expansion of
 Prove that
[8+8]
Solution
Now RR1: (2n + 1)xP_{n}(x) = (n + 1)P_{n+1}(x) + nP_{n−1}(x). Multiply by P_{n−1} (x) and integrate w.r.t x from − to 1
by Eq. (1)
3.
 Determine the analytic function f(z) = u + iv given that 3u + 2v = y^{2} − x^{2} + 16x
 If sin (α + iβ) = x + iy, then
[8 + 8]
Solution
 3u + 2v = y^{2} − x^{2} + 16x (1)
Differentiating w.r.t.x and y, we get
3u_{x} + 2v_{x} = −2x + 16 (2)
3u_{y} + 2v_{y} = −2y
⇒ 1u_{x} − 3v_{x} = 2y (by CREs) (3)
Eliminating v_{x} and u_{x} from Eqs. (2) and Eqs. (3), we
by MlineThomson’s rule of replacing x by z and y by 0
Integrating w.r.t.z, we have
(c is an arbitrary complex constant)
 x + = iy = sin(α + iβ)
= sin α cos hβ + i cos α sin hβ (1)
⇒ x = sin α cosh β, y = cos α sin hβ (2)
From Eq. (2), we have
Squaring and adding
Squaring and subtracting
4.
 Evaluate where using Cauchy’s integral formula
 State & prove Cauchy’s theorem.
Solution
 f(z) = log z is analytic within C and the singular point of the integrand is a = 1 (pole of order 3) lines inside C. By Cauchy’s integral formula,
 Please refer Theorem 6.1 in Ch. 6.
 Expand log z by Taylor’s series about z = 1
 Expand in positive and negative powers of z if 1
[8+8]
Solution
 Put z = 1 = w ⇒ z = 1 + w
∴ log z log (1 + w)
 Let
Putting into partial fractions
which is the required Laurent’s expansion
6.
 Determine the poles and the corresponding residues of
 Evaluate where C is the circle z = 4 using residue theorem.
[8+8]
Solution
 The poles of f(z) are at z = 0 (double pole) and z = 2 (simple pole)
 Let f(z)
The poles of f(z) are given by
sinh z = 0 ⇒ z = ±iπ (n : integer, since sin hz = sin iz)
7.
 Evaluate by residue theorem
 Use the method of contour intergration to evaluate
[8+8]
Solution
 Please refer to Ex. 6.20 in Ch. 6; Ch. 6 or Sol. to Qn. 7(a) Nov. 2006 Set 2.
 Please refer to the Solution of 7(b) in JNTU Nov. 2006 (Set 2) (Take a = π).
8. (a) (b) Please refer to Question 8 (a), (b) and its Solution in JNTU Nov. 2006 (Set 3).
Set No. 1 
Time: 3 hours 
Max Marks 80 
1.
 Show that
 Show that
Solution
 Please refer to Section 1.7 in Ch. 1.
 Please refer to Subsection 1.6.1 in Ch. 1.
2.
 Find the polar form of CauchyRiemann equations.
 Find whether is analytic or not.
Solution
 Please refer to Subsection 2.2.8 in Ch. 2.
 Let
Suppose (x, y) ≠ (0, 0)
CREs are satisfied at all points except at the origin ⇒ f(z) ∈ H(D−{0}).
3.
 Separate the real and imaginary parts of cot z.
 Prove that e^{z} is analytic.
 Find e^{z} and e^{z} if z = 4π (2 + i)
Solution
4.
 State and prove Cauchy’s integral theorem.
 Prove that
 and
Solution
 Please refer to Cauchy−Goursat theorem in Ch. 4.
 (i) and (ii) Please refer to Subsections 4.5.1 and 4.5.2 in Ch. 4.
5.
 Obtain the Laurent’s series expansion of .
 Expand in the region 1 < z < 2.
Solution
6.
 State and prove Cauchy’s Residue theorem.
 Find the residue of z = 0 of the function
Solution
 Please refer to Subsection 6.1.5 in Ch. 6.
 Res of f(z) at z = 0 is
7.
 State and prove the fundamental theorem of Algebra.
 Show that one root of the equation z^{4} + z^{3} + 1= 0 lies in the first quadrant.
Solution
 Please refer to Section 7.7 in Ch. 7.
 Please refer to Ex. 7.10 in Ch. 7.
8. (a) Under the transformation w = 1/z find the image of the circle z − 2i = 2.
(b) Find and plot the image of the triangular region with vertices at (0, 0) (1, 0) and (0, 1) under the transformation w = (1−i)z + 3.
Solution
 Please refer to Ex. 8.11 in Ch. 8.
 Let z = x + iy, w = u + iv; O = (0, 0), A = (1, 0), B = (0, 1) be the vertices of the triangle in the zplane. Suppose P, Q and R the images of O, A and B respectively in the wplane.
w = u + iv = (1 − i)(x + iy) + 3
= (x + y + 3)+i(− x + y)
(x, y) = (0, 0) ⇒ (u, v) = (3, 0)
(x, y) = (1, 0) ⇒ (u, v) = (4, −1)
(x, y) = (0, 1) ⇒ (u, v) = (4, 1)
Set No. 2 
Time: 3 hours 
Max Marks 80 
1.
 Show that .
 Prove that .
Solution
 Please refer to Subsection 1.4.4 in Ch. 1.
 Please refer to Section 1.9 in Ch. 1.
2.
 Show that z^{n} is analytic and hence find its derivative.
 Prove that is a harmonic function and find its harmonic conjugate.
Solution
 By the definition of the derivative
Hence exists for all z and is equal to nz^{n−1}.
 Some information is missing.
3.
 Separate the real and imaginary parts of sinh (x +iy).
 Find the general value of
 log(1 + );
 log (−1).
Solution
 Please refer to “Formulas of Hyperbolic Functions” in Ch. 3.
 Please refer to Ex. 3.12 (i) and (iii) in Ch. 3.
4.
 Evaluate aloing the path y = x and y = x^{2}.
 Evaluate using Cauchy’s integral formula where C is the circle.
Solution
 Please refer to Ex. 4.15 in Ch. 4.
 Incomplete. Please refer to Ex. 4.49 in Ch. 4. Here Cauchy’s integral theorem (not integral formula) is applicable.
5.
 Obtain Taylor’s expansion of about z = 2.
 State and prove Lauret’s theorem.
 Put z − 2 = t ⇒ z = t + 2. Now
 Please refer to Section 5.8 in Ch .5.
6. Prove that
where a > b > 0.
Solution
Please refer to Ex. 6.25 in Ch. 6.
7. State and prove Rouche’s theorem.
Solution
Please refer to Theorem 7.3 in Ch. 7.
8.
 State that the transformation w = e^{z} transforms the region between the real axis and the line parallel to the real axis at y = π into upper half of the wplane
 Find the bilinear transformation which maps z = −1, i, 1 into the points w = −i, 0, i.
Solution
 Please refer to Subsection 8.4.3 in Ch. 8.
 Please refer to Ex. 8.21 in Ch. 8 (interchange z and w).
Set No. 3 
Time: 3 hours 
Max Marks 80 
1.
 Show that
 Prove that
 Prove that
Solution
 Please refer to Subsection 1.3.2 in Ch. 1.
 Please refer to Subsection 1.4.1 in Ch. 1.
 Please refer to Subsection 1.6.5 in Ch. 1.
2.
 Find the analytic function whose real part is u = e^{2x} (xcos 2y − ysin2y).
 Find whether of f (z) = sin x sin y − i cos x cos y is analytic or not.
Solution
 Let f(z) = u + iv be the analytic function where
u = e^{2x} (xcos2y − ysin2y) (1)
Differentiating (1) w.r.t x and y
u_{x} = 2e^{2x}(xcos2y − ysin2y) + e^{2x} cos2y
u_{y} = e^{2x}(−xsin2y − sin2y − 2y cos2y)
We know that
f′(z) = u_{x} + iv_{x} = u_{x} − iu_{y} ∵ v_{x} = − u_{y}
= e^{2x} (2x cos2y − 2y sin2y + cos2y) + ie^{2x} (x sin2y + sin2y + 2y cos2y)
Replace x by z and y by 0 to obtain f′(z) by Milne Thomson’s method
f′(z) = e^{2z} (2z · 1 + 0 + 1) + ie^{2z} (0 + 0 + 0)
= (2z + 1)e^{2z}
Integrating we get
which is the required analytic function.
 Let f (z) = u + iv = sin x sin y − i cos x cos y
⇒ u = sin x sin y v = −cos x cos y
we have
u_{x} = cos x sin y u_{y} = sin x cos y
v_{x} = sin x cos y v_{y} = cos x sin y
CREs are u_{x} = v_{y} and u_{y} = −v_{x}. Clearly, the second equation is not satisfied. Hence f(z) is not analytic.
3.
 (Find the modulus and argument of i^{log(1+i)}.
 If tan (x + iy) = A + iB prove that A^{2} + B^{2}− 2B coth 2y + 1 = 0.
Solution
 Let x + iy = i^{log(1+i)} = e^{(1+i)log}, by def.
Let o = r cosθ, 1 = rsinθ
⇒ r = 1, θ =
= log 1 + i + 2nπi
Principle value = log 1 + i
Now
Modulus =
Argument =
(to make the denominator real)
∵ sin it = i sinh t, cos it = cosht;
4.
 Evaluate along y = x^{2}.
 Evaluate using Cauchy’s integral formula
Solution
5.
 Expand in a series of positive and negative power of z.
 Expand e^{z} as Taylor’s series about z = 1.
Solution
6
 Find the residues of
 Show that
Solution
 Singularities of f (z)
Pole of order 4 at z = 1 and simple poles are at
Let
where
by Formula IV in Ch. 6
Now
 Please refer to Ex. 6.20 in Ch. 6.
7. State and prove argument principle.
Solution
Please refer to Section 7.3 in Ch. 7.
8. Define conformal mapping. State and prove sufficient conditions for w = f (z) to represent a conformal mapping.
Solution
Please refer to Sections 8.1 and 8.2 in Ch. 8.
Code No. 43088 R07 
Set No. 4 
Time: 3 hours 
Max Marks 80 
1.
 Evaluate
 Show that
Solution
Put x = sin^{2} θ, dx = 2sinθ cosθ dθ
x = 0 ⇒ θ = 0, x = 1 ⇒ θ =
(a)
(ii)x = a sin θ, dx = a cosθ dθ
x = 0 ⇒ θ = 0, x = a ⇒ θ =
(iii) Put x^{2} = 8sin^{2}θ, 3x^{2}dx = 16sinθ cosθ dθ
(8 − x^{3})^{1/3} = (8cos^{2} θ)^{1/3} = 2cos ^{2/3} θ
x = 0 ⇒ θ = 0;
x = a ⇒ θ = π / 2
(b) Please refer to Ex. 1.10 in Ch. 1.
2.
 Prove that the function f(z) = is not analytic at the origin, although the CR equations are satisfied.
 Show that an analytic function of constant modulus is constant.
Solution
 Let f (z) = u (x, y) + i v (x, y) where u = and v = 0. Then at the origin
Similarly
Hence at the origin C.R.Es u_{x} = v_{y}, u_{y} = − v_{x} are satisfied.
Now
Let z = (x, y) → (0, 0) along y = mx, we get
The limit depends upon m and hence it is not unique.
∴ f ′(0) does not exist.
u_{x}, v_{x}, u_{y}, v_{x} all exist at (0, 0) and CREs are satisfied.
Since the partial derivatives are not continuous at (0, 0), f ′ (0) does not exist.
 Let f (z) ∈ H (D) and f (z) = K in D;
 f (z) ⇒ K u^{2} + v^{2} = K^{2}
Differentiating we get
uu_{x} + vv_{x} = 0 (1)Similarly
uu_{y} + vv_{y} = 0 (2)Using CREs u_{x} = v_{y}; u_{y} = − v_{x} we get
uu_{x} − vu_{y} = 0 (3)uu_{y} + vu_{x} = 0 (4)Eliminating u_{y} between Eqs. (3) and (4) we get
(u^{2} + v^{2}) u_{x} = 0and eliminating u_{x} between Eqs. (3) and (4)
(u^{2} + v^{2}) u_{y} = 0If k^{2} = u^{2} + v^{2} = 0 then u = 0, v = 0 ⇒ f(z) = 0
If k ≠ 0 then u_{x} = u_{y} = 0 ⇒ v_{x} = v_{y} = 0 by CREs
∴ u, v are constant ⇒ f(z) is constant.
3.
 Separate the real and imaginary parts of tanh z.
 Find all the roots of tanh z + 2 = 0.
Solution
4.
 Evaluate z^{2} dz along the line x = 2y.
 Use Cauchy’s integral formula to evaluate where C is the circle z =3.
Solution
 we can put into partial fractions as
The singularities a = −1, −2 lie inside C
By Cauchy’s generalized formula
Now I = I_{1} − I_{2} + I_{3} where
5
 Find the Taylor’s series to represent the function in the region z < 2.
 Expand f(z) = about (i) z = −1 and (ii) z = 1.
Solution
(a) (i) Put z + 1 = t ⇒ z = t − 1
(b) (ii) Put z = t + 1
6
 Find the poles of f (z) = and residues at the poles.
 Evaluate where C: z − i = 2.
Solution
 f(z) has simple poles at z = −1 and z = −3
 By Cauchy’s generalized integral formula
Here a = 2i, n = 1
The integrand double pole of z = xi which lies inside C.
C: x^{2} + (y − 2)^{2} = 4
7. Prove that all the roots of
 16z^{5} − z + 8 = 0 lie between the circles
 z^{6} − 9z^{2} + 11 = 0 lie between the circles z =1 and z =3.
Solution
 Let f(z) = 16z^{5}, g (z) = −z + 8 f, g ∈ H (z ≤1)
Also, f (z) has 5 zeros
By Rouche’s theorem,
f (z) = 16z^{5} and f (z) + g (z) = 16z^{5} − z + 8 have the same number of zeros inside the circle z =1. But f (z) has 5 zeros.
Hence 16z^{5} − z + 8 has 5 zeros within the circle z =1.
Again f (z) = 8, g (z) = 16z^{5} − z and f (z) has no zero
By Rouche’s theorem,
f (z) = 8 and f (z) + g (z) = 16z^{5} − z + 8 have the same number of zeros inside the circle z = i.e. no zeros inside z =
∴ All the 5 zeros of 16z^{5} − z + 8 = 0 lie between the circles z = and z = 1.
 Let f(z) = z^{6}, g(z) = − 9z^{2} + 11
where f, g ∈ H (z ≤3)
f (z) = z^{6} has 6 zeros.
By Rouche’s theorem,
f (z) and f (z) + g (z) = z^{6} − 9z^{2} + 11 have the same number of zeros inside the circle z = 3 i.e. 6 zeros inside z = 3
Also, let f (z) = 11, g (z) = z^{6} − 9z^{2} f, g ∈ H (z ≤ 1)
has no zeros inside z =1
By Rouche’s theorem,
f (z) = 11, f (z) + g (z) = z^{6} − 9z^{2} + 11 have the same number of zeros inside the circle z =1 i.e. no zeros inside z = 1
∴ All the 6 zeros of z^{6} − 9z^{2} + 11 lie between the circles z =1 and z =3.
8
 Find the image of the region in the z plane between the lines y = 0 and y = under transform w = e^{z}.
 Find the bilinear transformation which maps the points (− 1, 0, 1) into the points (0, i, 3i).
Solution
 Please refer to Subsection 8.4.3 in Ch. 8.
 Let the required bilinear transformation be
Here w_{1} = 0; w_{2} = i; w_{3} = 3i
z_{1} = −1; z_{2} = 0; z_{3} = 1
Eq. (1) now becomes
By componendodividendo