Cauchy-Riemann Equations Questions and Answers

This set of Complex Function Theory Multiple Choice Questions & Answers (MCQs) focuses on “Cauchy-Riemann (C-R) Equations”.

1. The function f(z) = xy + iy is continuous everywhere but not differential anywhere.
a) True
b) False
View Answer

Answer: a
Explanation:
Given f(z) = xy +iy
i.e. u=xy, v=y
x and y are continuous everywhere and consequently u(x,y) = xy and v(x,y) = y are continuous everywhere.
Thus, f(z) is continuous everywhere.
But

u=xy	v=y
u_x=y	v_x=0
u_y=x	v_y=1
u_x≠v_y	u_y≠u_x

C-R equations are not satisfied. Hence, f(z) is not differentiable anywhere
though it is continuous everywhere.

2. Which of the following is the derivative of the function f(z) = zⁿ?
a) nz^n-1
b) 1
c) 0
d) n²
View Answer

Answer:a
Explanation:
Let z= re^iθ
zⁿ=rⁿ(e^inθ)
zⁿ=rⁿ(cos ⁡nθ + isin⁡nθ)
i.e., u= rⁿcos⁡nθ, v= rⁿsin⁡nθ
\( \frac{∂u}{∂r}=nr^{n-1}cos⁡nθ, \frac{∂v}{∂r} = nr^{n-1}sin⁡nθ \)
\( \frac{∂u}{∂θ}=-nr^nsinnθ, \frac{∂v}{∂θ}=nr^ncosnθ \)
\( \frac{∂u}{∂r}=\frac{1}{r}\frac{∂v}{∂θ}, \frac{∂v}{∂r}=-\frac{1}{r}\frac{∂u}{∂θ}\)
C-R equations are satisfied and the partial derivatives are continuous.
Hence,the function is analytic.
\( f'(z)=\frac{u_r+iv_r}{e^{iθ}} = \frac{nr^{n-1}cos⁡nθ+in^{r-1}sin⁡nθ}{e^{iθ}} \)
\( =\frac{nr^ncos⁡nθ+inr^nsin⁡nθ}{re^{iθ}} = \frac{n[r(cosnθ+isinθ)]^n}{re^{iθ}} \)
\( = \frac{n[re^{iθ}]^n}{re^{iθ}} = \frac{nz^n}{z} = nz^{n-1} \)

3. Which of the following is the derivative of the function f(z) = 1/z?
a) \( \frac{-1}{z^2} \)
b) \( \frac{1}{z^3} \)
c) \( \frac{-1}{z^3} \)
d) \( \frac{1}{z^2} \)
View Answer

Answer: a
Explanation:
Let z=re^iθ
\( f(z) = \frac{1}{z}=\frac{1}{re^{iθ}}=\frac{1}{r}e^{-iθ} \)
\( \frac{1}{r}[cosθ-isinθ] \)
At z=0,r=0 and so f(z) is not defined at z=0
C-R equations are
\( \frac{∂u}{∂r}=\frac{1}{r}\frac{∂v}{∂θ}, \frac{∂u}{∂θ}=-r\frac{∂v}{∂r} \)
\( u=\frac{cosθ}{r} v=\frac{-sinθ}{r} \)
\( u_r=\frac{-cosθ}{r^2} v=\frac{sinθ}{r^2} \)
\( u_θ=\frac{-sinθ}{r} v_θ=\frac{-cosθ}{r} \)
C-R equations are satisfied.
Hence,f(z)is analytic everywhere except z=0.
f(z)=u+iv
Differentiating w.r.t. r’,
\( e^{iθ} f'(z) = u_r + iv_r \)
\( f'(z) = \frac{u_r+iv_r}{e^{iθ}}=\frac{-cosθ+isinθ}{r^2e^{iθ}}=\frac{-[cosθ-isinθ]}{r^2e^{iθ}} \)
\( =\frac{-e^{iθ}}{r^2e^{iθ}}=\frac{-1}{(re^iθ)^2}=\frac{-1}{z^2} \)

4. Which of the following co-ordinates is correct on which the function f(z)= √(|xy|) is not regular, although the Cauchy – Riemann equations are satisfied?
a) (0, 0)
b) (1, 2)
c) (-1, -2)
d) (2, 1)
View Answer

Answer: a
Explanation:
\(f(z)= u(x,y)+i v(x,y)= \sqrt{|xy|} \)
\( i.e., u(x,y)= \sqrt{|xy|}, v(x,y)=0 \)
\( u_x = (\frac{∂u}{∂x})_{(0,0)} = lim_{∆x→0}\frac{⁡0-0}{∆y} = 0 \)
\( u_y (0,0)= lim_{∆y→0}\frac{0-0}{∆y}=0 \)
\( v_x (0,0)= lim_{∆x→0}⁡\frac{0-0}{∆x}=0 \)
\( v_y (0,0)= lim_{∆y→0}⁡\frac{0-0}{∆y}=0 \)
Clearly, u_x = v_y and u_y = -v_x at the origin
\( Now, lim_{∆z→0}\frac{f(0+∆z)-f(0)}{∆z}= lim_{∆z→0}⁡\frac{\sqrt{|∆x∆y|}-0}{∆x+i∆y} \)
\( =lim_{∆y}= m∆x\frac{\sqrt{|m||∆x|^2}}{∆x(1+im)} = \frac{\sqrt{|m|}}{1+im} \)

The limit is not unique, since it depends on m. Therefore, f'(0) does not exist.
Hence, f(z) is not regular at the origin.

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5. Which of the following is not an analytic function?
a) 2xy + i(x²-y²)
b) xy
c) x + y
d) 2xy
View Answer

Answer: a
Explanation:
Let f(z) = 2xy + i(x-y²)

u=2xy	v=x²-y²
\(\frac{∂u}{∂x}=2y\)	\(\frac{∂v}{∂x}=2x\)
\(\frac{∂u}{∂y}=2x\)	\(\frac{∂v}{∂y}=-2y\)

u_x≠v_y and u_y≠-v_x
Hence, C-R equations are not satisfied.
Hence, f(z) is not an analytic function.

6. Which of the following co-ordinates is correct on which the function f(z) = x²+iy³ is not regular, although the Cauchy-Riemann equations are satisfied?
a) (0,0)
b) (1, 2)
c) (2,1)
d) (-1,-2)
View Answer

Answer: a
Explanation:
Let f(z) = x³ +iy³
i.e., u+iv=x²+iy³

u=x²	v=y³
u_x=2x	v_x=0
u_y=0	v_y=3y²

u_x≠v_y and u_y= -v_x
C-R equations are not satisfied.
Hence, f(z) is not analytic.
2x=3y² is true at x=0, y=0
Hence, C-R equations are satisfied at (0,0)

7. Which of the following curves is correct on which the function f(z)= (x-y)²+ 2i (x+y) is analytic and C-R equations are satisfied?
a) x-y=1
b) x+y=1
c) x²+y²=1
d) x²-y²=1
View Answer

Answer: a
Explanation:
Given: f(z)=(x-y)²+ 2i(x+y)
i.e., u+iv=(x-y)²+ i2(x+y)

u=(x-y)²	v=2(x+y)
u_x=2(x-y)	v_x=2
u_y=2(x-y)(-1)=-2(x-y)	v_y=2

u_x= v_y, u_y=-v_x
2(x-y)= 2, -2(x-y)= -2
x-y=1, x-y=1
Hence, C-R equations are satisfied only if x-y=1.

8. Which of the following is the derivative of f(z)=cos ⁡hz?
a) sin ⁡hz
b) cosec hz
c) sec⁡ hz
d) tan ⁡hz
View Answer

Answer: a
Explanation:
\( Given: f(z)= \cos(⁡hz) = \cos⁡(iz)=cos⁡([i(x+iy)]) \)
\( =\cos⁡(ix-y)=\cos⁡(ix)\cos⁡(y)+\sin⁡(ix)\sin(⁡y) \)
\( u+iv=\cos⁡(hx)\cos⁡(y)+i\sin⁡(hx)\sin(⁡y) \)

\(u=\cos(hx)cos(y)\)	\(v=\sin(hx)sin(y)\)
\(u_x=\sin(hx)cos(y)\)	\(v_x=\cos(hx)sin(y)\)
\(u_y=-\cos(hx)sin(y)\)	\(v_y=\sin(hx)cos(y)\)

u_x,u_y, v_x,v_y exist and are continuous.
u_x=v_y and u_y=-v_x
C-R equations are satisfied
f(z)is analytic everywhere.
\( Now, f'(z) = u_x+iv_x = \sin(hx)\cos(y) + i\cos(hx)\sin(y) = \sin(h(x+y)) =\sin(hz) \)

9. Which of the following is analytic if u+iv is also analytic?
a) v-iu and-v+iu
b) u+iv and u-iv
c) v+iu and v-iu
d) -v+iu and u+iv
View Answer

Answer: a
Explanation:
Given: u+iv is analytic
C-R equations are satisfied.
i.e., u_x= v_y and u_y= -v_x
Since the derivatives of u and v exist therefore it is continuous.
Now, to prove v-iu and -v+iu are also analytic, we should prove that
(i) v_x= -u_y and v_y= u_x and
(ii) -v_x=u_y and v_y= u_x
(iii) u_x,u_y,v_x and v_y are continuous. Since the derivatives of u and v exists from (1) and (2),the derivatives of u and v should be continuous.
Hence, the result (iii) follows.

10. Which of the following is true ifw=f(z) and z=x+iy?
a) \( \frac{dw}{dz}= \frac{∂w}{∂x}=-i\frac{∂w}{∂y} \)
b) \( \frac{dw}{dz}= \frac{∂w}{∂x}= i\frac{∂w}{∂y} \)
c) \( \frac{dw}{dz}= \frac{∂w}{∂x}= \frac{∂w}{∂y} \)
d) \( \frac{dw}{dz}= \frac{∂w}{∂x}= -2\frac{∂w}{∂y} \)
View Answer

Answer: a
Explanation:
Let w=u(x,y)+ iv(x,y)
As f(z) is analytic, we have u_x=v_y,u_y= -v_x
\( Now, \frac{dw}{dz}=f'(z)= u_x+iv_x= v_y-iu_y=-i(u_y+iv_y) \)
\( = \frac{∂u}{∂x}+i\frac{∂v}{∂x}= -i[\frac{∂u}{∂y}+i\frac{∂v}{∂y}] \)
\( = \frac{∂}{∂x}(u+iv)= -i\frac{∂}{∂y}(u+iv) \)
\( = \frac{∂w}{∂x}=-i\frac{∂w}{∂y} \)

11. Which of the following are the values of a, b, c if f(z)=(x+ay)+i(bx+cy) is analytic?
a) a=b and c=0
b) a= -b and c=1
c) a=-b and c=0
d) a= -b and c=0
View Answer

Answer: b
Explanation:
If f(z)= u(x,y)+ iv(x,y) is analytic, then f(z)= u(z,0)+ iv(z,0)
Hence, f(z)= z+ibz [x=z and y=0] =(x+iy)+ib(x+iy)
=x+ibx+iy-by
Given: f(z)= x+ay+i(bx+c)
i.e., x+ay+i(bx+cy) = x+ibx+iy-by
By Equating, we get
a=-b, c=1

12. For what values of z, the function w defined by z= e^(-v)(cos ⁡u + i sin ⁡u) ceases to be analytic?
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: a
Explanation:
The function w=f(z) will cease to be analytic at points, where \( \frac{dz}{dw}=0 \)
\(z=e^{-v} e^{iu}= e^{i(u+iv)} \)
\( z= e^{iw} \)
\(\frac{dz}{dw}=ie^{iw}\) =iz = 0 when z=0
Thus, w ceases to be analytic at z=0.

13. For what values of z, the function w defined by \( \frac{z}{(z^2-1)}\) ceases to be analytic?
a) ±1
b) 0
c) 2
d) -2
View Answer

Answer: a
Explanation:
\(Let f(z)= \frac{z}{z^2-1} \)
\(f'(z)= \frac{(z^2-1)(1)- z(2z)}{(z^2-1)^2} = \frac{-(z^2+1)}{(z^2-1)^2} \)
f(z) is not analytic, where f^’ (z)does not exist
i.e.,f’ (z)→∞
i.e., (z²-1)²=0
i.e.,z²-1=0
z= ±1

14. For what values of z, the function w defined by \(\frac{z+i}{(z-i)^2}\) ceases to be analytic?
a) i
b) 0
c) 1
d) 2
View Answer

Answer: a
Explanation:
\(Let f(z)= \frac{z+i}{(z-i)^2} \)
\(f'(z)= \frac{(z-i)^2(1)-(z+i)[2(z-i)]}{(z-i)^4} = \frac{-(z+3i)}{(z-i)^3} \)
f'(z)→ ∞, at z=i
f(z)is not analytic at z=i

15. For what values of z, the function w defined by tan²z ceases to be analytic?
a) \(\frac{(2n-1)π}{2}\),n=1,2,3,…
b) nπ,n=1,2,3,…
c) (n-1)π,n=1,2,3,…
d) 2nπ,n=1,2,3,…
View Answer

Answer: a
Explanation:
\(Let f(z)= \tan^2(z) \)
\(f'(z)=2\tan⁡(z)\sec^2(z)= \frac{2sin⁡z}{\cos^3(z)} \)
\(f'(z)→∞, at \cos^3(z)=0 \)
\(i.e., z=\frac{(2n-1)}{2}π \)
∴f(z) is not analytic at \(z= \frac{(2n-1)π}{2},n=1,2,3,… \)

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