# Question Papers – Engineering Mathematics, Volume III

## Question Papers

 Code No.R059210201 Set No.1

II Year B.Tech. I Semester Supplementary Examinations, February 2008

Engineering Mathematics-III

 Time: 3 hours Max Marks: 80

Answer any FIVE Questions
All questions carry equal marks

1.

1. Evaluate dx using βγ functions
2. Evaluate using βγ functions
3. Evaluate dx using B, Г functions

[5+6+5]

Solution

1. Put log

The limits for t are ∞, 0.

2. Put

The limits for t are 0, ∞.

Note:

The upper limit should be ‘a’ and not −1 as printed.

3. Put x = a sinθ, dx = a cosθ

The limits for θ are 0, π/2.

2.

1. Show that

Solution

1. Putting n = 1, 2, 3, we get from Eq. (1)

Substituting for J2(x) from Eq. (2) into Eq. (3), we get

Substituting for J2(x) from Eq. (5) and J2(x) from Eq. (2) into Eq. (4) we get

2. Pn(x) is a solution of Legendre’s equation

(1 − x2)y″ − 2xy′ + n(n + 1)y = 0 or

[(1 − x2)y′]′ + n(n + 1)y = 0                         (1)

n(n + 1)Pn(x) = −[(1 − x2)P′n (x)]′         (2)

Multiplying both sides by Pn(x) and integrating by parts w.r.t x from − 1 to 1

3.

1. If w = u + iv is an analytic function of z and
2. If sin(θ + ) = cos α + i sin α, then prove that cos4 θ = sin2 α.

[8+8]

Solution

1. Please refer to Ex. 2.45 in Ch. 2.
2. sin(θ + ) = 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

⇒ cos2 α − cos2 θ − sin2 α sin2 θ = sin2 θ cos2 θ

⇒ (1 − sin2 α) cos2 θ − sin2 α(1 − cos2 θ)

= (1 − cos2 θ) cos2 θ ⇒ cos4 θ = sin2 α.

4.

1. Find f(2) and f(3) if f(a) = where C is the circle |z| = 2.5 using Cauchy’s integral formula
2. Evaluate where C is the circle |z| = 1 using Cauchy’s integral formula.

Solution

1. 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.

2. Put x = e log z = log e = ; dz = ie

The limits for θ are 0 to 2π

5.

1. Find the Laurent expansion of for 1 < |z| < 3
2. Expand in a Laurent series for |z| < 3.

[8+8]

Solution

1. Putting into partial fraction, we have for 1 < |z| < 3 the Laurent series expansion is

6.

1. Find the poles and the residues at the poles of
2. Evaluate where C is |z| = 5 by residue theorem.

[8+8]

Solution

1. The poles of f(z) = are the zeros of the denominator z2 + 1 = 0

x = i, −i, which are simple poles of f(z)

2. The poles of f(z) = are the zeros of the denominator z2 + 9 = 0, i.e., z = 3i, −3i which are simple poles of f(z) lying inside |z| = 5

By Cauchy’s residue theorm,

7.

1. Show that (a2 < 1) using Residue Theorem
2. Show by the method of contour integration that

Solution

1. Please refer Ex. 6.28 in Ch. 6.
2. Let C be the closed contour consisting of
1. Line Segment L [-R,R] along the real axis
2. Semicircle SR : z = Re (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.

1. Show that image of the hyperbola x2y2 = 1 under the transformation w = 1/z is r2 = cos 2θ
2. Show that the transformation changes the circle x2 + y2 − 4x = 0.

[8+8]

Solution

1. Let z = re, w = Re. The transformation becomes

The hyperbola x2y2 = 1 becomes

r2 cos2 θr2 sin2 θ = 1

r2(cos2 θ − sin2 θ) = 1 or r2 cos2θ = 1

cos(−2ϕ) = 1 by Eq. (1)

R2 = cos2ϕ

Thus the hyperbola x2y2 = 1 under the transformation becomes lemniscate R2 = cos 2ϕ.

2. Solving for z, we have

Now the equation of the circle x2 + y2 − 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

II Year B.Tech. I Semester Supplementary Examinations, February 2008

Engineering Mathematics-III

 Time: 3 hours Max Marks: 80

Answer any FIVE Questions

All questions carry equal marks

1.

1. Show that β(m, n) = γ(m)γ(n)/γ(m + n)
2. Show that where n is an odd integer
3. Show that

Solution

1. Please refer Theorem 6.1 in Ch. 6.
2. Put x = sin θ, dx = cos θdθ, The limits for θ are 0 and π/2.

since n is odd put n = 2m + 1

3.

2.

1. Establish the formula Pn+1(x) − Pn−1(x) = (2n + 1)Pn(x)
2. Prove that
3. When n is an integer, show that J−n(x) = (−1)nJn(x).

[6+5+5]

Solution

1. Please refer to RR1 (Legendre polynomisdl) Ch. 1.
2. Please refer to RR2 (Bessel functions) Ch. 1.
3. Please refer to the proof of Ex. 1.57(i) in Ch. 1 (Bessel functions.)

3.

1. Find the analytic function whose imaginary part is f(x, y) = x3yxy3 + xy + x + y where z = x + iy
2. Prove that where w = f(z) is analytic.

[8+8]

Solution

1. Let

ϕ(z) = g(x, y) + if (x, y) ∈ H(D)                 (1)

whose imaginary part is

f(x, y) = x3yxy3 + xy + x + y             (2)

f is harmonic function in Dfxx + fyy = 0. we have,

fx = 3x2yy3 + y + 1 and

fy = x3 − 3xy2 + x + 1

fxx = 6xy and fyy = − 6xy

⇒ fxx + fyy = 0                             (3)

Also, f, g satisfy CREs:

gx = fy and gy = − fx                     (4)

by Milne-Thomson’s rule of replacing x by z and y by 0.

Integrating Eq. (8) w.r.t z, we get

2. Let f(z) = u + ivH(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 JNTU-Regular Exam Nov. 2008 (Set 1).

5.

1. State and prove Laurent’s theorem.
2. Obtain all the Laurent series of the function about z = −2.

[8+8]

Solution

1. Please refer to Ch. 5.
2. Please refer to Ex. 5.30 in Ch. 5.

6.

1. Find the poles and residue at each pole of
2. Find the poles and residue at each pole of
3. Evaluate where C is |z| = 3/2.

Solution

1. The poles of are the zeros of the denominator
2. 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
3. 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.

1. Show that (ab ≥0) using residue theorem.
2. Evaluate, by contour integration,

[8+8]

Solution

1. 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 (αβ) of which a lies inside C while lies outside C.

By Cauchy’s residue theorem,

2. Consider where C consists of the line segment [-R, R] along the real axis and the semicircle CR : |z| = R in the upper half-plane. 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 CR: z = Re, bz = CR: z = Re and limits for θ are 0, π. Therefore we have,

8.

1. Show that the transformation maps the interior of the circle |z| = 1 into the upper-half of the w-plane, the upper semicircle into the positive half of the real axis and the lower semicircle into the negative half of the real axis.
2. By the transformation w = z2, show that the circle |za| = c (a and c are real) in the z-plane corresponds to the binacon in the w-plane.

Solution

1. 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 w-plane.

Also |z| = 1 (the unit circle) can be defined by z = e (0 ≤0 2π)

The transformation (1) gives

on multiplying the numerator and denominator by 2eeiθ/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 z-plane)

u ≤ 0, v = 0 for + π/2 ≤ θ, 2 ≤ π or π ≤ θ ≤ 2 π (lower semicircle in the z-plane)

Thus, the transformation (1) maps the upper semicircle in the z-plane into the positive half of the real axis (u ≥ 0, v = 0) and the lower semicircle in the z-plane into the negative half of the real axis (u ≤ 0, v = 0).

2. The equation of the circle |za| = c may be written as za = cez = a + ce

The mapping w = z2 becomes

By taking the pole (origin) in the w-plane at w = a2c2, we put w − (a2c2) = Re we have

Re = 2ce(a + c cos θ)

Equating the modulus and amplitude of the complex quantity on either side

R = 2c(a + c cos θ) and ϕ = 0

so that R = 2c(a + ccos ϕ). This equation represents a binacon in the w-plane. Thus the circle |za| = c in the z-plane corresponds to binacon in the w-plane.

 Code No.R059210201 Set No.3

II Year B.Tech. I Semester Supplementary Examinations, February 2008

Engineering Mathematics-III

 Time: 3 hours Max Marks: 80

Answer any FIVE Questions

All questions carry equal marks

1.

1. Show that
2. Prove that
3. Show that

[5+5+6]

Solution

1. By definition,

Put x = sin2 θdx = 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

From Eqs. (4) and (5) the result follows.

2. We know that,

Taking q = 1 − p we get from (5) and (6)

From calculus of residues, we have the result

From (7) and (8) the result follows.

3.

2.

1. Prove that

[16]

Solution Please refer to the orthogonal property of Legendre polynomials in Ch. 1

3

1. Find the analytic function whose imaginary part is f(x,y) = x3yxy3 + xy + x + y where z = x + iy
2. Prove that where f(z) is analytic

[8+8]

Solution

1. Let f(z) = u+iv where v = x3yxy3+xy+x+y Differentiating v w.r.t x and y, we get

f′(z) = ux + ivx = vy + ivx

= (x3 − 3xy2 + x + 1) + i(3x2yy3 + y + 1)

(by Cauchy-Riemann Equations)

= (z3 − 0 + z + 1) + i(0 − 0 + 0 + 1),

replacing x by z and y by 0,
according to Milne Thomson’s method

= z3 + z + 1 + i

Integrating w.rtz, f(z) = 1/4z4 + 1/2z2 + z + iz + c, where c is a complex constant, is the required function.

2. We have

4.

1. Evaluate with C : |z| = 2 using cauchy’s integral formula.
2. Evaluate
3. Evaluate where C : |z| = 2 using Cauchy’s integral theorem.

[5+5+6]

Solution

1. f(z) = (z3 − sin 3z) is analytic inside the circle C : |z| = 2 and the singular point a = π/2 of the integrand lies inside C.
2.

C is the circle |z| = 2. The singularity z = 1 lies inside C.f(z) = e−z is analytic f″(z) = e−zf″(1) = e−1.

5.

1. Expand about z = 1 as a Laurent series. Also find the region of convergence.
2. Find Taylor series for about z = 1; also find the region of convergence.

Solution

1. Put z − 1 = w or z = w + 1. Then

which is the region of convergence.

2. 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.

1. Find the poles and the residue at each pole of
2. Evaluate where C is |z| = 5, by residue theorem.

[8+8]

Solution

1. Please refer to the Solution of 6(a) in JNTU Feb 2008 (Set 1).
2. Please refer to Solution of 6(b) in JNTU Feb 2008 (Set 1).

7.

1. Evaluate using residue theorem.
2. 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 (0 ≤ θ ≥ 2π),

Poles of the integrand are , −2 of which lies inside C

(b) We evaluate the integral

by considering the contour integral Cf(z)dz where f(z) = and C is the closed path consisting of the semicircel CR : z = Re (0 ≤ θπ) 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 (z2 + 1) (z2 + 4), of which z = i, 2i lie within C in the upper half of the plane.

By residue theorem,

Now on CR : z = Re (0 ≤ θπ) dz = iRe

Letting R → ∞ we obtain from Eqs. (5) and Eqs. 6

8.

1. Find the image of the infinite strip 0 < y < under the transformation w = 1 /z
2. Find the bilinear transformation which maps the points (−1, 0, 1) into the points (0, i, 3i).

[8+8]

Solution

1. 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 u2 + (v + 1)2 = 1. Thus the infinite strip 0 < y < in the z-plane is mapped into the region between the line v = 0 and the circle u2 + (v + 1)2 = 1 under the mapping w =

2. Let the transformation be

Substituting in Eq. (1), we get

by componendo-dividendo

which is the required transformation.

 Code No. 2059210201 Set No.4

II Year B.Tech. I Semester Supplementary Examinations, February 2008

Engineering Mathematics III

 Time: 3 hours Max Marks: 80

Answer any FIVE Questions

All questions carry equal marks

1.

1. Evaluate using βγ
2. Prove that using βγ function and evaluate.
3. Show that

[5+5+6]

Solution

1.

2.

[by putting x2 = tx = dt and the limits for t are 0, ∞]

[by putting x4 = tx = dt and the limits for t are 0, ∞]

From Eqs. (1) and (2), we have

3. Put x = at, dx = adt and the limits for t are 0, ∞.

2.

1. Show that the coefficient of tn in the power series expansion of
2. Prove that

[8+8]

Solution

1. See Ex. 1.46 in Chapter 1
2. By orthogonality of Legendre polynomials, we have

Now RR1: (2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x). Multiply by Pn−1 (x) and integrate w.r.t x from − to 1

by Eq. (1)

3.

1. Determine the analytic function f(z) = u + iv given that 3u + 2v = y2x2 + 16x
2. If sin (α + ) = x + iy, then

[8 + 8]

Solution

1. 3u + 2v = y2x2 + 16x             (1)

Differentiating w.r.t.x and y, we get

3ux + 2vx = −2x + 16                         (2)

3uy + 2vy = −2y

⇒ 1ux − 3vx = 2y (by CREs)                     (3)

Eliminating vx and ux from Eqs. (2) and Eqs. (3), we

by Mline-Thomson’s rule of replacing x by z and y by 0

Integrating w.r.t.z, we have

(c is an arbitrary complex constant)

2. x + = iy = sin(α + )

= sin α cos + i cos α sin         (1)

⇒ x = sin α cosh β, y = cos α sin         (2)

From Eq. (2), we have

Squaring and adding

Squaring and subtracting

4.

1. Evaluate where using Cauchy’s integral formula
2. State & prove Cauchy’s theorem.

Solution

1. 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,
2. Please refer Theorem 6.1 in Ch. 6.

5.

1. Expand log z by Taylor’s series about z = 1
2. Expand in positive and negative powers of z if 1

[8+8]

Solution

1. Put z = 1 = wz = 1 + w

∴ log z log (1 + w)

2. Let

Putting into partial fractions

which is the required Laurent’s expansion

6.

1. Determine the poles and the corresponding residues of
2. Evaluate where C is the circle |z| = 4 using residue theorem.

[8+8]

Solution

1. The poles of f(z) are at z = 0 (double pole) and z = 2 (simple pole)
2. Let f(z)

The poles of f(z) are given by

sinh z = 0 ⇒ z = ± (n : integer, since sin hz = sin iz)

7.

1. Evaluate by residue theorem
2. Use the method of contour intergration to evaluate

[8+8]

Solution

1. Please refer to Ex. 6.20 in Ch. 6; Ch. 6 or Sol. to Qn. 7(a) Nov. 2006 Set 2.
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).

 Code No. 43088 R07 Set No. 1

II Year B.Tech. I Semester Regular Examinations November 2009

Engineering Mathematics − III

(Common to EEE, ECE, EIE, E CON. E, ETM, ICE)

 Time: 3 hours Max Marks 80

Answer any FIVE questions

All questions carry equal marks

1.

1. Show that
2. Show that

Solution

1. Please refer to Section 1.7 in Ch. 1.
2. Please refer to Subsection 1.6.1 in Ch. 1.

2.

1. Find the polar form of Cauchy-Riemann equations.
2. Find whether is analytic or not.

Solution

1. Please refer to Subsection 2.2.8 in Ch. 2.
2. Let

Suppose (x, y) ≠ (0, 0)

CREs are satisfied at all points except at the origin ⇒ f(z) ∈ H(D−{0}).

3.

1. Separate the real and imaginary parts of cot z.
2. Prove that ez is analytic.
3. Find ez and |ez| if z = 4π (2 + i)

Solution

1. Please refer to Ex. 3.8 in Ch. 3.
2. Please refer to definition and properties of ez in Ch. 3.
3.

4.

1. State and prove Cauchy’s integral theorem.
2. Prove that
1. and

Solution

1. Please refer to Cauchy−Goursat theorem in Ch. 4.
2. (i) and (ii) Please refer to Subsections 4.5.1 and 4.5.2 in Ch. 4.

5.

1. Obtain the Laurent’s series expansion of .
2. Expand in the region 1 < |z| < 2.

Solution

1. Put z − 1 = tz = 1 + t.

Then

2. Please refer to Ex. 34 Ch. 5.

6.

1. State and prove Cauchy’s Residue theorem.
2. Find the residue of z = 0 of the function

Solution

1. Please refer to Subsection 6.1.5 in Ch. 6.
2. Res of f(z) at z = 0 is

7.

1. State and prove the fundamental theorem of Algebra.
2. Show that one root of the equation z4 + z3 + 1= 0 lies in the first quadrant.

Solution

1. Please refer to Section 7.7 in Ch. 7.
2. 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

1. Please refer to Ex. 8.11 in Ch. 8.
2. Let z = x + iy, w = u + iv; O = (0, 0), A = (1, 0), B = (0, 1) be the vertices of the triangle in the z-plane. Suppose P, Q and R the images of O, A and B respectively in the w-plane.

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)

 Code No. 43088 R07 Set No. 2

II Year B.Tech. I Semester Regular Examinations November 2009

Engineering Mathematics − III

(Common to EEE, ECE, EIE, E CON. E, ETM, ICE)

 Time: 3 hours Max Marks 80

Answer any FIVE questions

All questions carry equal marks

1.

1. Show that .
2. Prove that .

Solution

1. Please refer to Subsection 1.4.4 in Ch. 1.
2. Please refer to Section 1.9 in Ch. 1.

2.

1. Show that zn is analytic and hence find its derivative.
2. Prove that is a harmonic function and find its harmonic conjugate.

Solution

1. By the definition of the derivative

Hence exists for all z and is equal to nzn−1.

2. Some information is missing.

3.

1. Separate the real and imaginary parts of sinh (x +iy).
2. Find the general value of
1. log(1 + );
2. log (−1).

Solution

1. Please refer to “Formulas of Hyperbolic Functions” in Ch. 3.
2. Please refer to Ex. 3.12 (i) and (iii) in Ch. 3.

4.

1. Evaluate aloing the path y = x and y = x2.
2. Evaluate using Cauchy’s integral formula where C is the circle.

Solution

1. Please refer to Ex. 4.15 in Ch. 4.
2. Incomplete. Please refer to Ex. 4.49 in Ch. 4. Here Cauchy’s integral theorem (not integral formula) is applicable.

5.

1. Obtain Taylor’s expansion of about z = 2.
2. State and prove Lauret’s theorem.

Solution

1. Put z − 2 = tz = t + 2. Now
2. 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.

1. State that the transformation w = ez transforms the region between the real axis and the line parallel to the real axis at y = π into upper half of the w-plane
2. Find the bilinear transformation which maps z = −1, i, 1 into the points w = −i, 0, i.

Solution

1. Please refer to Subsection 8.4.3 in Ch. 8.
2. Please refer to Ex. 8.21 in Ch. 8 (interchange z and w).

 Code No. 43088 R07 Set No. 3

II Year B.Tech. I Semester Regular Examinations November 2009

Engineering Mathematics − III

(Common to EEE, ECE, EIE, E CON. E, ETM, ICE)

 Time: 3 hours Max Marks 80

Answer any FIVE questions

All questions carry equal marks

1.

1. Show that
2. Prove that
3. Prove that

Solution

1. Please refer to Subsection 1.3.2 in Ch. 1.
2. Please refer to Subsection 1.4.1 in Ch. 1.
3. Please refer to Subsection 1.6.5 in Ch. 1.

2.

1. Find the analytic function whose real part is u = e2x (xcos 2yysin2y).
2. Find whether of f (z) = sin x sin yi cos x cos y is analytic or not.

Solution

1. Let f(z) = u + iv be the analytic function where

u = e2x (xcos2yysin2y) (1)

Differentiating (1) w.r.t x and y

ux = 2e2x(xcos2yysin2y) + e2x cos2y

uy = e2x(−xsin2y − sin2y − 2y cos2y)

We know that

f′(z) = ux + ivx = uxiuyvx = − uy

= e2x (2x cos2y − 2y sin2y + cos2y) + ie2x (x sin2y + sin2y + 2y cos2y)

Replace x by z and y by 0 to obtain f′(z) by Milne Thomson’s method

f′(z) = e2z (2z · 1 + 0 + 1) + ie2z (0 + 0 + 0)

= (2z + 1)e2z

Integrating we get

which is the required analytic function.

2. Let f (z) = u + iv = sin x sin yi cos x cos y

u = sin x sin y v = −cos x cos y

we have

ux = cos x sin y uy = sin x cos y

vx = sin x cos y vy = cos x sin y

CREs are ux = vy and uy = −vx. Clearly, the second equation is not satisfied. Hence f(z) is not analytic.

3.

1. (Find the modulus and argument of ilog(1+i).
2. If tan (x + iy) = A + iB prove that A2 + B2− 2B coth 2y + 1 = 0.

Solution

1. Let x + iy = ilog(1+i) = e(1+i)log, by def.

Let o = r cosθ, 1 = rsinθ

r = 1, θ =

log i = log r + + i(2)

= log 1 + i + 2nπi

Principle value = log 1 + i

Now

Modulus =

Argument =

2.

(to make the denominator real)

∵ sin it = i sinh t, cos it = cosht;

4.

1. Evaluate along y = x2.
2. Evaluate using Cauchy’s integral formula

Solution

1. z = x + iy = x + ix2dz = (1 + 2xi)dx
2. Please refer to Ex. 4.45 in Ch. 4.

5.

1. Expand in a series of positive and negative power of z.
2. Expand ez as Taylor’s series about z = 1.

Solution

1.

2.

6

1. Find the residues of
2. Show that

Solution

1. 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

2. 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

II Year B.Tech. I Semester Regular Examinations November 2009

Engineering Mathematics − III

(Common to EEE, ECE, EIE, E CON. E, ETM, ICE)

 Time: 3 hours Max Marks 80

Answer any FIVE questions

All questions carry equal marks

1.

1. Evaluate
2. Show that

Solution

Put x = sin2 θ, dx = 2sinθ cosθ

x = 0 ⇒ θ = 0, x = 1 ⇒ θ =

(a)

(ii)x = a sin θ, dx = a cosθ

x = 0 ⇒ θ = 0, x = aθ =

(iii) Put x2 = 8sin2θ, 3x2dx = 16sinθ cosθ

(8 − x3)1/3 = (8cos2 θ)1/3 = 2cos 2/3 θ

x = 0 ⇒ θ = 0;

x = aθ = π / 2

(b) Please refer to Ex. 1.10 in Ch. 1.

2.

1. Prove that the function f(z) = is not analytic at the origin, although the CR equations are satisfied.
2. Show that an analytic function of constant modulus is constant.

Solution

1. 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 ux = vy, uy = − vx 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.

ux, vx, uy, vx all exist at (0, 0) and CREs are satisfied.

Since the partial derivatives are not continuous at (0, 0), f ′ (0) does not exist.

2. Let f (z) ∈ H (D) and |f (z)| = K in D;

| f (z)| ⇒ K u2 + v2 = K2

Differentiating we get

uux + vvx = 0                 (1)

Similarly

uuy + vvy = 0                 (2)

Using CREs ux = vy; uy = − vx we get

uuxvuy = 0                 (3)

uuy + vux = 0                 (4)

Eliminating uy between Eqs. (3) and (4) we get

(u2 + v2) ux = 0

and eliminating ux between Eqs. (3) and (4)

(u2 + v2) uy = 0

If k2 = u2 + v2 = 0 then u = 0, v = 0 ⇒ f(z) = 0

If k ≠ 0 then ux = uy = 0 ⇒ vx = vy = 0 by CREs

u, v are constant ⇒ f(z) is constant.

3.

1. Separate the real and imaginary parts of tanh z.
2. Find all the roots of tanh z + 2 = 0.

Solution

1. Please refer to Ex. 3.9 in Ch. 3.
2.

By componendo-dividendo

4.

1. Evaluate z2 dz along the line x = 2y.
2. Use Cauchy’s integral formula to evaluate where C is the circle |z| =3.

Solution

1.

2. we can put into partial fractions as

The singularities a = −1, −2 lie inside C

By Cauchy’s generalized formula

Now I = I1I2 + I3 where

5

1. Find the Taylor’s series to represent the function in the region |z| < 2.
2. Expand f(z) = about (i) z = −1 and (ii) z = 1.

Solution

(a) (i) Put z + 1 = tz = t − 1

(b) (ii) Put z = t + 1

6

1. Find the poles of f (z) = and residues at the poles.
2. Evaluate where C: |z − i| = 2.

Solution

1. f(z) has simple poles at z = −1 and z = −3
2. By Cauchy’s generalized integral formula

Here a = 2i, n = 1

The integrand double pole of z = xi which lies inside C.

C: x2 + (y − 2)2 = 4

7. Prove that all the roots of

1. 16z5z + 8 = 0 lie between the circles
2. z6 − 9z2 + 11 = 0 lie between the circles |z| =1 and |z| =3.

Solution

1. Let f(z) = 16z5, g (z) = −z + 8 f, g H (|z| ≤1)

Also, f (z) has 5 zeros

By Rouche’s theorem,

f (z) = 16z5 and f (z) + g (z) = 16z5z + 8 have the same number of zeros inside the circle |z| =1. But f (z) has 5 zeros.

Hence 16z5z + 8 has 5 zeros within the circle |z| =1.

Again f (z) = 8, g (z) = 16z5z and f (z) has no zero

By Rouche’s theorem,

f (z) = 8 and f (z) + g (z) = 16z5z + 8 have the same number of zeros inside the circle |z| = i.e. no zeros inside |z| =

∴ All the 5 zeros of 16z5z + 8 = 0 lie between the circles |z| = and |z| = 1.

2. Let f(z) = z6, g(z) = − 9z2 + 11

where f, gH (z ≤3)

f (z) = z6 has 6 zeros.

By Rouche’s theorem,

f (z) and f (z) + g (z) = z6 − 9z2 + 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) = z6 − 9z2 f, g H (z ≤ 1)

has no zeros inside |z| =1

By Rouche’s theorem,

f (z) = 11, f (z) + g (z) = z6 − 9z2 + 11 have the same number of zeros inside the circle |z| =1 i.e. no zeros inside |z| = 1

∴ All the 6 zeros of z6 − 9z2 + 11 lie between the circles |z| =1 and |z| =3.

8

1. Find the image of the region in the z plane between the lines y = 0 and y = under transform w = ez.
2. Find the bilinear transformation which maps the points (− 1, 0, 1) into the points (0, i, 3i).

Solution

1. Please refer to Subsection 8.4.3 in Ch. 8.
2. Let the required bilinear transformation be

Here w1 = 0; w2 = i; w3 = 3i

z1 = −1; z2 = 0; z3 = 1

Eq. (1) now becomes

By componendo-dividendo

which is the required bilinear transformation.