# Complex Function Theory Questions and Answers – Analyticity

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 = ex sin⁡y?
a) -iez + C
b) iez + C
c) ez + C
d) -ez + C

Explanation:
Given: u = ex sin⁡y
ux = ex sin⁡y, ux (z,0) = 0     [∴ sin⁡0 = 0]
uy = ex cos⁡y, uy (z,0) = ez     [∴ cos⁡0 = 1]
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy     [by C.R.equation]
f(z) = ∫ ux (z,0)dz – i∫ uy (z,0) dz + C     [by Milne – Thomson rule]
where C is a complex constant.
∴ f(z) = ∫0 dz – i∫ez dz + C
= -iez + C

2. Which of the following is the analytic function f(z) for which the real part is ex cos⁡y?
a) iez + C
b) -iez + C
c) ez + C
d) -ez + C

Explanation:
Given: u = ex cos⁡y
ux = ex cos⁡y, ux (z,0) = ez     [∴ cos⁡0 = 1]
uy = -ex sin⁡y, uy (z,0) = 0     [∴sin⁡0 = 0]
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy     [by C.R. condition]
f(z) = ∫ ux (z,0) dz – i∫ uy (z,0) dz + C [by Milne – Thomson rule] where C is a complex constant.
∴ f(z) = ∫ez dz – i∫0 dz + C
= ez + C

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

Explanation:
Given: u = y + ex cos⁡y
ux = 0 + ex cos⁡y
uy = 1 – ex sin⁡y
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy     [by C.P.condition]
f(z) = ∫ ux (z,0)dz – i∫ uy (z,0) dz + C     [by Milne – Thomson rule]
where C is a complex constant
f(z) = ∫ez dz – i∫dz + C
= ez – iz + C

4. Which of the following is the analytic function w = u + iv if u = e2x (x cos⁡2y – y sin⁡2y) ?
a) ze2z + C
b) -ze2z + C
c) e2z + C
d) z + C

Explanation:
Given: u = e2x (x cos⁡2y – y sin⁡2y)
ux = e2x [cos⁡2y] + (x cos⁡2y – y sin⁡2y)
ux (z,0) = e2z [1] + [z(1) – 0][2e2z] = e2z + 2z e2z
= (1 + 2z) e2z i.e., ux (z,0) = (1 + 2z)e2z
uy = e2x [- 2x sin⁡2y – (y2 cos⁡2y + sin⁡2y)] uy (z,0) = e2z [- 0 – (0 + 0) ] = 0 i.e., uy (z,0) = 0
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy
f(z) = ∫ ux (z,0) dz – i ∫ uy (z,0)dz + C [by Milne – Thomson rule] where C is a complex constant.
f(z) = ∫(1 + 2z) e2z dz – i∫0 dz + C
= ∫(1 + 2z) e2z dz + C
= (1 + 2z) e2z/2 – 2e2z/4 + C     [∴ ∫uv dz = uv1 – u v2 + u v3 – …]
= $$\frac {e^{2z}}{2}$$ + ze2z – $$\frac {e^{2z}}{2}$$ + C = ze2z + 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

Explanation:
Given: u = x3 – 3xy2 + 3x2 – 3y2 + 1
ux = 3x2 – 3y2 + 6x, ux (z,0) = 3z2 – 0 + 6z
uy = 0 – 6xy + 0 – 6y, uy (z,0) = 0
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy     [by C.R.Equation]
f(z) = ∫ ux (z,0)dz – i∫ uy (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

Explanation:
u = $$\frac {sin⁡2x}{cosh⁡2y – cos⁡2x}$$

ux = $$\frac {cosh⁡2y – cos⁡2x)[2 cos⁡2x] – sin⁡2x [2 sin⁡2x]}{[cosh⁡2y – cos⁡2x] ^{2 }}$$
ux (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., ux (z,0) = -cosec2z
uy = $$\frac {(cosh⁡2y – cos⁡2x)(0) – sin⁡2x[2 sinh⁡2y]}{cosh⁡2y – cos⁡2x) ^{2}}$$

uy (z,0) = 0
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy     [by C.R.condition]
f(z) = ∫ ux (z,0)dz – i∫ uy (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) zez + C
b) -zez + C
c) ez + C
d) -ez + C

Explanation:
u = ex (x cos⁡y – y sin⁡y)
ux = ex [cos⁡y] + (x cos⁡y – y sin⁡y) [ex]
ux (z,0) = ez (1) + (z – 0) ez = ez + zez = (1 + z) ez
uy = ex [- xsin y – (y cos⁡y + sin⁡y)]
uy (z,0) = ez [- 0 – (0 + 0) ] = 0
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy     [by C.R.condition]
f(z) = ∫ ux (z,0)dz – i∫ uy (z,0) dz + C     [by Milne – Thomson method]
f(z) = ∫(1 + z) ez dz – i(0) + C
= (1 + z) ez dz – (1) ez + C
f(z) = zez + 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

Explanation:
Given: u = $$\frac {1}{2}$$ log⁡(x2 + y2)
ux = $$\frac {1}{2} \frac {1}{x^2 + y^{2}}$$(2x) = $$\frac {x}{x^2 + y^{2′}}$$, ux (z,0) = $$\frac {1}{z}$$
uy = latex]\frac {1}{2} \frac {1}{x^2 + y^{2}}[/latex](2x) = $$\frac {x}{x^2 + y^{2′}}$$, uy (z,0) = 0
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy     [by C.R.condition]
f(z) = ∫ ux (z,0)dz – i∫ uy (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 = ex2 – y2 cos⁡2xy?
a) ez2 + C
b) -ez2 + C
c) 1 + ez2 + C
d) 1 – ez2 + C

Explanation:
u = ex2 – y2 cos⁡2xy = ex2 e-y2 cos⁡2xy
ux = e-y2 [ex2(- 2y sin⁡2xy) + cos⁡2xy ex22x] ux (z,0) = 2zez2
uy (z,0) = ez2 (0 + 0) = 0
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy     [by C.R.condition]
f(z) = ∫ ux (z,0)dz – i∫ uy (z,0) dz + C [by Milne – Thomson method] where C is a complex constant
= ∫2 z ex2 dz + C
Put t = z2, dt = 2z dz
= ∫et dt + C
= et + C
f(z) = ex2 + 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

Explanation:
v = e-x (x cos⁡y + y sin⁡y)
vx = e-x [cos⁡y] + (x cos⁡y + y sin⁡y)[- e-x]
vx (z,0) = e-z + (z)( – e-z) = (1 – z) e-z vx (z,0) = (1 – z)e-z
vy = e-z [- x sin⁡y + (y cos⁡y + sin⁡y (1))] vy (z,0) = e-z [0 + 0 + 0] = 0
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy     [by C.R.condition]
f(z) = ∫ ux (z,0)dz – i∫ uy (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

Explanation:
Given: ∅ = tan-1⁡($$\frac {y}{x}$$)
We should denote, φ by u and ∅ by v
∴v = tan-1⁡($$\frac {y}{x}$$)
vx = $$\frac {1}{1 + (\frac {y}{x})^{2}} \big [ \frac { – y}{x^{2}} \big ] = \frac { – y}{x^2 + y^{2′}}$$, vx (z,0) = 0
vy = $$\frac {1}{1 + (\frac {y}{x})^{2}} \big [ \frac { 1}{x} \big ] = \frac { x}{x^2 + y^{2′}}$$, vy (z,0) = $$\frac {1}{z}$$
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy [by C.R.condition] f(z) = ∫ ux (z,0)dz – i∫ uy (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

Explanation:
Given: v = is x2 – y2 + $$\frac {x}{x^2 + y^{2}}$$
vx = 2x – 0 + $$\frac {(x^2 + y^2)(1) – x(2x)}{(x^2 + y^2)^{2}}$$
= 2x + $$\frac {y^2 – x^{2}}{x^2 + y^{2})^{2}}$$, vx (z,0) = 2z + $$\frac {-z^{2}}{(z^{2})^{2}}$$
vx (z,0) = 2z – $$\frac {1}{z^{2}}$$
vy = 0 – 2y + $$\frac {0 – x(2y)}{x^2 + y^2){2}}$$
= -2y $$\frac {- 2xy}{(x^2 + y^2)^{2}}$$, vy (z,0) = 0
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy [by C.R.condition]
f(z) = ∫ ux (z,0)dz – i∫ uy (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

Explanation:
Let f(z) = e-2x cos⁡2y
fx = – 2e-2x cos⁡2y
fyyxx = 4e-2x cos⁡2y
fy = – 2 e-2x sin⁡2y
fyy = – 4 e-2x cos⁡2y
f_xx + fyy = 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 = ex (cos⁡y – sin⁡y) ?
a) ez + C
b) – ez + C
c) iez + C
d) – iez + C

Explanation:
Given: u – v = ex (cos⁡y – sin⁡y) ………(1)
Differentiating (1) w.r.t. x,we get
ux – vx = ex (cos⁡y – sin⁡y)
ux (z,0) – vx (z,0) = ez………(2)
Differentiating (1) w.r.t. y,we get
uy – vy = ex [- sin⁡y – cos⁡y]
uy (z,0) – vy (z,0) = ez (- 1)
– vx (z,0) – ux (z,0) = – ez………(3) [by C.R.condition]
(1) + (2) = – 2 vx (z,0) = ez – ez = 0
vx (z,0) = 0
ux (z,0) = ez
Let w = f(z) = u + iv
f(z) = ux + i vx
= ux – i uy [by C.R.condition]
f(z) = ∫ ux (z,0)dz – i∫ uy (z,0) dz + C [by Milne – Thomson method]
where C is a complex constant
f(z) = ∫ez dz + i(0) + C
= ez + C

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

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)
vx = 2(1 – y)
vx x = 0
vy = -2x
vyy = 0
Thus, vx x + vyy = 0
Hence, v is harmonic and therefore, it can be the imaginary part of an analytic function.

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