Analyticity - Complex Function Theory Questions and Answers

This set of Complex Function Theory Multiple Choice Questions & Answers (MCQs) focuses on “Analyticity”.

1. Which of the following is the function such that w = u + iv is analytic, ifu = e^x sin⁡y?
a) -ie^z + C
b) ie^z + C
c) e^z + C
d) -e^z + C
View Answer

Answer: a
Explanation:
Given: u = e^x sin⁡y
u_x = e^x sin⁡y, u_x (z,0) = 0 [∴ sin⁡0 = 0]
u_y = e^x cos⁡y, u_y (z,0) = e^z [∴ cos⁡0 = 1]
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.R.equation]
f(z) = ∫ u_x (z,0)dz – i∫ u_y (z,0) dz + C [by Milne – Thomson rule]
where C is a complex constant.
∴ f(z) = ∫0 dz – i∫e^z dz + C
= -ie^z + C

2. Which of the following is the analytic function f(z) for which the real part is e^x cos⁡y?
a) ie^z + C
b) -ie^z + C
c) e^z + C
d) -e^z + C
View Answer

Answer: c
Explanation:
Given: u = e^x cos⁡y
u_x = e^x cos⁡y, u_x (z,0) = e^z [∴ cos⁡0 = 1]
u_y = -e^x sin⁡y, u_y (z,0) = 0 [∴sin⁡0 = 0]
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.R. condition]
f(z) = ∫ u_x (z,0) dz – i∫ u_y (z,0) dz + C [by Milne – Thomson rule] where C is a complex constant.
∴ f(z) = ∫e^z dz – i∫0 dz + C
= e^z + C

3. Which of the following is the analytic function f(z) = u(x,y) + iv(x,y) whose real part is y + e^x cos⁡y?
a) e^z – iz + C
b) e^z + iz + C
c) -e^z – iz + C
d) -e^z + iz + C
View Answer

Answer: a
Explanation:
Given: u = y + e^x cos⁡y
u_x = 0 + e^x cos⁡y
u_y = 1 – e^x sin⁡y
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.P.condition]
f(z) = ∫ u_x (z,0)dz – i∫ u_y (z,0) dz + C [by Milne – Thomson rule]
where C is a complex constant
f(z) = ∫e^z dz – i∫dz + C
= e^z – iz + C

4. Which of the following is the analytic function w = u + iv if u = e^2x (x cos⁡2y – y sin⁡2y) ?
a) ze^2z + C
b) -ze^2z + C
c) e^2z + C
d) z + C
View Answer

Answer: a
Explanation:
Given: u = e^2x (x cos⁡2y – y sin⁡2y)
u_x = e^2x [cos⁡2y] + (x cos⁡2y – y sin⁡2y)
u_x (z,0) = e^2z [1] + [z(1) – 0][2e^2z] = e^2z + 2z e^2z
= (1 + 2z) e^2z i.e., u_x (z,0) = (1 + 2z)e^2z
u_y = e^2x [- 2x sin⁡2y – (y2 cos⁡2y + sin⁡2y)] u_y (z,0) = e^2z [- 0 – (0 + 0) ] = 0 i.e., u_y (z,0) = 0
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y
f(z) = ∫ u_x (z,0) dz – i ∫ u_y (z,0)dz + C [by Milne – Thomson rule] where C is a complex constant.
f(z) = ∫(1 + 2z) e^2z dz – i∫0 dz + C
= ∫(1 + 2z) e^2z dz + C
= (1 + 2z) e^2z/2 – 2e^2z/4 + C [∴ ∫uv dz = uv₁ – u^‘ v₂ + u^” v₃ – …]
= \(\frac {e^{2z}}{2}\) + ze^2z – \(\frac {e^{2z}}{2}\) + C = ze^2z + C

5. Which of the following is the analytic function where real part is u = x³ – 3xy² + 3x² – 3y² + 1?
a) z³ + 3z² + C
b) -z³ + 3z² + C
c) z³ – 3z² + C
d) -z³ – 3z² + C
View Answer

Answer: a
Explanation:
Given: u = x³ – 3xy² + 3x² – 3y² + 1
u_x = 3x² – 3y² + 6x, u_x (z,0) = 3z² – 0 + 6z
u_y = 0 – 6xy + 0 – 6y, u_y (z,0) = 0
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.R.Equation]
f(z) = ∫ u_x (z,0)dz – i∫ u_y (z,0)dz + C [by Milne – Thomson method] where C is a complex constant
f(z) = ∫(3z² + 6z)dz – i∫0 dz + C
= 3 \(\frac {z^{3}}{3}\) + 6 \(\frac {z^{2}}{2}\) + C
= z³ + 3z² + C

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6. Which of the following is the analytic function whose real part is \(\frac {sin⁡2x}{cosh⁡2y – cos⁡2x}\)?
a) z³ + 3z² + C
b) z³ – 3z² + C
c) cot⁡z + C
d) -cot⁡z + C
View Answer

Answer: c
Explanation:
u = \(\frac {sin⁡2x}{cosh⁡2y – cos⁡2x}\)

u_x = \(\frac {cosh⁡2y – cos⁡2x)[2 cos⁡2x] – sin⁡2x [2 sin⁡2x]}{[cosh⁡2y – cos⁡2x] ^{2 }}\)
u_x (z,0) = \(\frac {(1 – cos⁡2z) (2 cos⁡2z) – 2sin^2 2z}{1 – cos⁡2z)^{2}}\)

= \(\frac {2 cos⁡2z – 2 cos^2 2z – 2sin^2 2z}{1 – cos⁡2z)^{2}} = \frac {2 cos⁡2z – 2[cos^2 2z + sin^2 2z]}{(1 – cos⁡2z)^{2}}\)

= \(\frac {2 cos⁡2z – 2}{(1 – cos⁡2z)^{2}} = \frac { – 2(1 – cos⁡2z)}{(1 – cos⁡2z)^{2}} = \frac {- 2}{1 – cos⁡2z} = \frac {- 2}{2sin^2 z}\)
= -cosec²z i.e., u_x (z,0) = -cosec²z
u_y = \(\frac {(cosh⁡2y – cos⁡2x)(0) – sin⁡2x[2 sinh⁡2y]}{cosh⁡2y – cos⁡2x) ^{2}}\)

u_y (z,0) = 0
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.R.condition]
f(z) = ∫ u_x (z,0)dz – i∫ u_y (z,0) dz + C [by Milne – Thomson method]
where C is a complex constant.
f(z) = ∫ – cosec²z dz – i ∫0 dz + C
= cot⁡z + C

7. Which of the following is the analytic function whose real part is (x cos⁡y – y sin⁡y) ?
a) ze^z + C
b) -ze^z + C
c) e^z + C
d) -e^z + C
View Answer

Answer: a
Explanation:
u = e^x (x cos⁡y – y sin⁡y)
u_x = e^x [cos⁡y] + (x cos⁡y – y sin⁡y) [e^x]
u_x (z,0) = e^z (1) + (z – 0) e^z = e^z + ze^z = (1 + z) e^z
u_y = e^x [- xsin y – (y cos⁡y + sin⁡y)]
u_y (z,0) = e^z [- 0 – (0 + 0) ] = 0
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.R.condition]
f(z) = ∫ u_x (z,0)dz – i∫ u_y (z,0) dz + C [by Milne – Thomson method]
f(z) = ∫(1 + z) e^z dz – i(0) + C
= (1 + z) e^z dz – (1) e^z + C
f(z) = ze^z + C

8. Which of the following is the analytic function whose real part is \(\frac {1}{2}\) log⁡(x² + y²)?
a) log⁡z + C
b) log \(\frac {1}{z}\) + C
c) log⁡z² + C
d) log⁡z³ + C
View Answer

Answer: a
Explanation:
Given: u = \(\frac {1}{2}\) log⁡(x² + y²)
u_x = \(\frac {1}{2} \frac {1}{x^2 + y^{2}}\)(2x) = \(\frac {x}{x^2 + y^{2′}}\), u_x (z,0) = \(\frac {1}{z}\)
u_y = latex]\frac {1}{2} \frac {1}{x^2 + y^{2}}[/latex](2x) = \(\frac {x}{x^2 + y^{2′}}\), u_y (z,0) = 0
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.R.condition]
f(z) = ∫ u_x (z,0)dz – i∫ u_y (z,0) dz + C [by Milne – Thomson method]
where C is a complex constant
f(z) = ∫\(\frac {1}{z}\) dz – i ∫0 dz + C
= log⁡z + C

9. Which of the following is an analytic function f(z) = u + iv, given that u = e^{x² – y²} cos⁡2xy?
a) e^z² + C
b) -e^z² + C
c) 1 + e^z² + C
d) 1 – e^z² + C
View Answer

Answer: a
Explanation:
u = e^{x² – y²} cos⁡2xy = e^x² e^-y² cos⁡2xy
u_x = e^-y² [e^x²(- 2y sin⁡2xy) + cos⁡2xy e^x²2x] u_x (z,0) = 2ze^z²
u_y (z,0) = e^z² (0 + 0) = 0
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.R.condition]
f(z) = ∫ u_x (z,0)dz – i∫ u_y (z,0) dz + C [by Milne – Thomson method] where C is a complex constant
= ∫2 z e^x² dz + C
Put t = z², dt = 2z dz
= ∫e^t dt + C
= e^t + C
f(z) = e^x² + C

10. Which of the following is the analytic function whose imaginary part is e^-x (x cos⁡y + y sin⁡y) ?
a) i z e^-z + C
b) -i z e^-z + C
c) z e^-z + C
d) i e^-z + C
View Answer

Answer: a
Explanation:
v = e^-x (x cos⁡y + y sin⁡y)
v_x = e^-x [cos⁡y] + (x cos⁡y + y sin⁡y)[- e^-x]
v_x (z,0) = e^-z + (z)( – e^-z) = (1 – z) e^-z v_x (z,0) = (1 – z)e^-z
v_y = e^-z [- x sin⁡y + (y cos⁡y + sin⁡y (1))] v_y (z,0) = e^-z [0 + 0 + 0] = 0
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.R.condition]
f(z) = ∫ u_x (z,0)dz – i∫ u_y (z,0) dz + C [by Milne – Thomson method]
where C is a complex constant
f(z) = ∫0 dz + i∫(1 – z) e^-z dz + C
= i∫(1 – z) e^-z dz + C
= i\( \bigg [ \)(1 – z)\(\big [\frac {e^{-z}}{ – 1} \big ]\) – ( – 1)\(\big [\frac {e^{-z}}{ – 1^{2}} \big ] \bigg ] \) + C
= i[- (1 – z) e^-z + e^-z ] + C
= i z e^-z + C

11. Which of the following is the velocity potential φ if the stream function is ∅ = tan^-1⁡(\(\frac {y}{x}\))?
a) log⁡z + C
b) log \(\frac {1}{z}\)⁡ + C
c) log z²⁡ + C
d) log⁡z³ + C
View Answer

Answer: a
Explanation:
Given: ∅ = tan^-1⁡(\(\frac {y}{x}\))
We should denote, φ by u and ∅ by v
∴v = tan^-1⁡(\(\frac {y}{x}\))
v_x = \(\frac {1}{1 + (\frac {y}{x})^{2}} \big [ \frac { – y}{x^{2}} \big ] = \frac { – y}{x^2 + y^{2′}}\), v_x (z,0) = 0
v_y = \(\frac {1}{1 + (\frac {y}{x})^{2}} \big [ \frac { 1}{x} \big ] = \frac { x}{x^2 + y^{2′}}\), v_y (z,0) = \(\frac {1}{z}\)
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.R.condition] f(z) = ∫ u_x (z,0)dz – i∫ u_y (z,0) dz + C [by Milne – Thomson method] where C is a complex constant
f(z) = ∫\(\frac {1}{z}\) dz + i ∫0 dz + C
= log⁡z + C
So, the velocity potential is ∅ = \(\frac {1}{2}\) log⁡(x² + y²)

12. Which of the following is the real part if the imaginary part is x² – y² + \(\frac {x}{x^2 + y^{2}}\)?
a) i[z² + \(\frac {1}{z}\) ] + C
b) – i[z² + \(\frac {1}{z}\) ] + C
c) [z² + \(\frac {1}{z}\) ] + C
d) – [z² + \(\frac {1}{z}\) ] + C
View Answer

Answer: a
Explanation:
Given: v = is x² – y² + \(\frac {x}{x^2 + y^{2}}\)
v_x = 2x – 0 + \(\frac {(x^2 + y^2)(1) – x(2x)}{(x^2 + y^2)^{2}}\)
= 2x + \(\frac {y^2 – x^{2}}{x^2 + y^{2})^{2}}\), v_x (z,0) = 2z + \(\frac {-z^{2}}{(z^{2})^{2}}\)
v_x (z,0) = 2z – \(\frac {1}{z^{2}}\)
v_y = 0 – 2y + \(\frac {0 – x(2y)}{x^2 + y^2){2}}\)
= -2y \(\frac {- 2xy}{(x^2 + y^2)^{2}}\), v_y (z,0) = 0
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.R.condition]
f(z) = ∫ u_x (z,0)dz – i∫ u_y (z,0) dz + C [by Milne – Thomson method]
where C is a complex constant
f(z) = ∫0 dz + i∫(2z – \(\frac {1}{z^{2}}\)) dz + C
= i\( \bigg [ \frac {(2z^{2}}{2} + \frac {1}{z} \bigg ] \) + C
= i[z² + \(\frac {1}{z}\) ] + C

13. The function f(z) = e^-2x cos⁡2y can be the real or imaginary part of an analytic function.
a) True
b) False
View Answer

Answer: a
Explanation:
Let f(z) = e^-2x cos⁡2y
f_x = – 2e^-2x cos⁡2y
f_yyxx = 4e^-2x cos⁡2y
f_y = – 2 e^-2x sin⁡2y
f_yy = – 4 e^-2x cos⁡2y
f_xx + f_yy = 0
In a simply connected domain, every harmonic function is the real part or the imaginary part
of some analytic function.
Therefore, Given function can be a real or imaginary part of of an analytic function.

14. Which of the following is an analytic function f(z) in terms of z if u – v = e^x (cos⁡y – sin⁡y) ?
a) e^z + C
b) – e^z + C
c) ie^z + C
d) – ie^z + C
View Answer

Answer: a
Explanation:
Given: u – v = e^x (cos⁡y – sin⁡y) ………(1)
Differentiating (1) w.r.t. x,we get
u_x – v_x = e^x (cos⁡y – sin⁡y)
u_x (z,0) – v_x (z,0) = e^z………(2)
Differentiating (1) w.r.t. y,we get
u_y – v_y = e^x [- sin⁡y – cos⁡y]
u_y (z,0) – v_y (z,0) = e^z (- 1)
– v_x (z,0) – u_x (z,0) = – e^z………(3) [by C.R.condition]
(1) + (2) = – 2 v_x (z,0) = e^z – e^z = 0
v_x (z,0) = 0
u_x (z,0) = e^z
Let w = f(z) = u + iv
f^‘(z) = u_x + i v_x
= u_x – i u_y [by C.R.condition]
f(z) = ∫ u_x (z,0)dz – i∫ u_y (z,0) dz + C [by Milne – Thomson method]
where C is a complex constant
f(z) = ∫e^z dz + i(0) + C
= e^z + C

15. 2x(1 – y) can be the imaginary part of an analytic function.
a) True
b) False
View Answer

Answer: a
Explanation:
In a simply connected domain, every harmonic function is the real part or imaginary part of some
analytic function.
Let v = 2x(1 – y)
v_x = 2(1 – y)
v_x x = 0
v_y = -2x
v_yy = 0
Thus, v_x x + v_yy = 0
Hence, v is harmonic and therefore, it can be the imaginary part of an analytic function.

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