Transformation w = sin z Questions and Answers

This set of Complex Mapping Multiple Choice Questions & Answers (MCQs) focuses on “Transformation w = sin z”.

1. Which of the following is the image in w plane when x = 0 under the transformation w=sin⁡z?
a) u=0
b) v=0
c) \(\frac{u^2}{sin^2 c}- \frac{v^2}{cos^2c}=1 \)
d) \(\frac{u^2}{cos⁡h^2c}+ \frac{v^2}{sin⁡h^2c}=1 \)
View Answer

Answer: a
Explanation:
Given:\( w= \sin{⁡z} \)
\(u+iv= \sin⁡{(x+iy)}\)
\(u+iv= \sin{x} \cos{⁡hy}+i \cos{x} \sin{⁡hy} \)
\(u=\sin⁡{x}\cos{⁡hy} …(1) \)
\(v=\cos⁡{x}\sin⁡{hy} …(2) \)
Given: x=0
Substituting x=0 in (1)
We get u=0.

2. Which of the following is the image in w plane when y = 0 under the transformation w=sin⁡z?
a) u=0
b) v=0
c) \(\frac{u^2}{sin^2 c}- \frac{v^2}{cos^2 c}=1 \)
d) \(\frac{u^2}{cos⁡h^2 c}+ \frac{v^2}{sin⁡h^2 c}=1 \)
View Answer

Answer: b
Explanation:
Given: \(w=\sin{⁡z} \)
\(u+iv=\sin⁡{(x+iy)} \)
\(u+iv=\sin{x} \cos{⁡hy}+i\cos{x} \sin{⁡hy} \)
\(u=\sin⁡{x}\cos{⁡hy}…(1) \)
\(v=\cos⁡{x} \sin{⁡hy} …(2) \)
Given: y=0
Substituting y=0 in (1)
We get v=0.

3. Which of the following is the image in w plane when y=c ≠0 under the transformation w=sin⁡z?
A) u=0
b) v=0
c) \(\frac{u^2}{sin^2 c}- \frac{v^2}{cos^2 c}=1 \)
d) \(\frac{u^2}{cos⁡h^2 c}+ \frac{v^2}{sin⁡h^2 c}=1 \)
View Answer

Answer: d
Explanation:
Given: \(w=\sin{⁡z} \)
\(u+iv=\sin⁡{(x+iy)} \)
\(u+iv=\sin{x} cos⁡{hy}+i \cos{x} \sin⁡{hy} \)
\(u=\sin⁡{x} \cos⁡{hy} …(1) \)
\(v=\cos⁡{x} \sin{⁡hy} …(2) \)
\(From (1)→\frac{u}{cos⁡hc} =\sin{⁡x} \)
\(From (2)→ \frac{v}{sinhc} =\cos{⁡x} \)
We know that \(\sin^2{x}+ \cos^2{x}=1 \)
Substituting the relevant terms,
\(\frac{u^2}{cos⁡h^2c}+ \frac{v^2}{sin⁡h^2c}=1\), which represents the equation of Ellipse.

4. Which of the following is the image in w plane when x=c ≠0 under the transformation w=sin⁡z?
a) u=0
b) v=0
c) \(\frac{u^2}{sin2 c}- \frac{v^2}{cos^2 c}=1 \)
d) \(\frac{u^2}{cos⁡h^2 c}+ \frac{v^2}{sin⁡h^2 c}=1 \)
View Answer

Answer: c
Explanation:
Given: \(w=\sin{⁡z} \)
\(u+iv=\sin{⁡(x+iy)} \)
\(u+iv=\sin{x} \cos⁡{hy}+i\cos{x} \sin{⁡hy} \)
\(u=\sin⁡{x} \cos⁡{hy} …(1) \)
\(v=\cos⁡{x} \sin{⁡hy} …(2) \)
From (1), \(\cos⁡{hy}= \frac{u}{sin⁡x} \)
From (2), \(\sin⁡{hy}= \frac{v}{cos⁡x} \)
We know that \(cosh^2 y- sinh^2 y=1 \)
Substituting the relevant terms,
\(\frac{u^2}{sin^2 x}- \frac{v^2}{cos^2 x}=1 \)
Given: x=c ≠0
\(\frac{u^2}{sin^2 c}- \frac{v^2}{cos^2 c}=1 \)

5. Which of the following is the image in w plane when x = 0 under the transformation w=cos⁡z?
a) \(u=\cos{⁡hy} \)
b) \(v=0 \)
c) \(\frac{u^2}{sin^2 c}- \frac{v^2}{cos^2 c}=1 \)
d) \(\frac{u^2}{cos⁡h^2 c}+ \frac{v^2}{sin⁡h^2 c}=1 \)
View Answer

Answer: a
Explanation:
Given: \( w=\cos{⁡z} \)
\(u+iv=\cos⁡{(x+iy)} \)
\(u+iv=\cos{x} \cos{hy}-\sin{x} \sin{hy} \)
\(u=\cos⁡{x} \cos{⁡hy} …(1) \)
\(v=-\sin⁡{x}\sin{⁡hy} …(2) \)
Given: x=0
Substituting x=0 in (1)
We get \(u= \cos{hy}. \)

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6. Which of the following is the image in w plane when y = 0 under the transformation w=cos⁡z?
a) \(u=\cos{⁡hy} \)
b) \(v=0 \)
c) \(\frac{u^2}{sin^2 c}- \frac{v^2}{cos^2 c}=1 \)
d) \(\frac{u^2}{cos⁡h^2 c}+ \frac{v^2}{sin⁡h^2 c}=1 \)
View Answer

Answer: b
Explanation:
Given: \(w=\cos{⁡z} \)
\(u+iv=\cos⁡{(x+iy)} \)
\(u+iv=\cos{x} \cos{hy}-\sin{x} \sin{hy} \)
\(u=\cos⁡{x} \cos{⁡hy} …(1) \)
\(v=-\sin{x} \sin{⁡hy} …(2) \)
Given: y=0
Substituting y=0 in (2)
We get v=0.

7. Which of the following is an image of the line \(x= -\frac{π}{2}\) under the mapping w=sinz?
a) u≤-1
b) u=0
c) u≥1
d) u≥-1
View Answer

Answer: a
Explanation:
Given: \(w=\sin{⁡z} \)
\(u+iv=\sin{⁡(x+iy)} \)
\(u+iv=\sin{x}\cos⁡{hy}+i\cos{x}\sin{⁡hy} \)
\(u=\sin⁡{x}\cos{⁡hy} …(1) \)
\(v=\cos⁡{x}\sin{⁡hy} …(2) \)
When \(x= -\frac{π}{2} \)
\(u=\sin⁡{(-π/2)}\cos{⁡hy} \)
\(u= -\cos{⁡hy} \)
\(v=\cos⁡{(π/2)} \sin{⁡hy} \)
v=0
We know that cos⁡hy≥1
-cos⁡hy ≤ -1
u ≤ -1
\(x=-\frac{π}{2}\) in z-plane is transformed into u≤1 in w-plane.

8. Which of the following is an image of the line \(x=-\frac{π}{2}\) under the mappin gw=sinz?
a) u≤-1
b) u=0
c) u≥1
d) u≥-1
View Answer

Answer: c
Explanation:
Given: \(w=\sin{⁡z} \)
\(u+iv=\sin⁡{(x+iy)} \)
\(u+iv=\sin{x} \cos⁡{hy}+i\cos{x} \sin{⁡hy} \)
\(u=\sin⁡{x} \cos{⁡hy} …(1) \)
\(v=\cos⁡{x} \sin⁡{hy} …(2) \)
When x= π/2
\(u=\sin⁡{(\frac{π}{2})}\cos{⁡hy} \)
\(u= \cos{⁡hy} \)
\(v=\cos⁡{(\frac{π}{2})} \sin{⁡hy} \)
v=0
We know that cos(⁡hy)≥1
cos⁡(hy) ≥1
u ≥1

9. Which of the following is an image of the line x= π under the mapping w=sinz?
a) u=0
b) u≥1
c) u≤1
d) u≥-1
View Answer

Answer: a
Explanation:
Given: w=sin⁡z
\(u+iv=\sin⁡{(x+iy)} \)
\(u+iv=\sin{x} \cos⁡{hy}+i \cos{x} \sin{⁡hy} \)
\(u=\sin⁡{x} \cos{⁡hy} …(1) \)
\(v=\cos⁡{x} \sin{⁡hy} …(2) \)
When x=π
u=sin⁡(π)cos(⁡hy)
u= 0

10. Which of the following is an image of the line x= -π under the mapping w=sinz?
a) u=0
b) u≥1
c) u≤1
d) u≥-1
View Answer

Answer: a
Explanation:
Given: w=sin⁡z
\(u+iv=\sin⁡{(x+iy)} \)
\(u+iv=\sin{x} \cos⁡{hy}+i\cos{x} \sin{⁡hy} \)
\(u=\sin⁡{x} \cos{⁡hy} …(1) \)
\(v=\cos⁡{x} \sin{⁡hy} …(2) \)
When x=-π
\(u=\sin⁡{(-π)}\cos{⁡hy} \)
u= 0

11. Which of the following is the image in w plane when x=π under the transformation w=cos⁡z?
a) v=0
b) v≥1
c) v≤-1
d) v≥-1
View Answer

Answer: c
Explanation:
Given: w=cos⁡z
\(u+iv=\cos⁡{(x+iy)} \)
\(u+iv=\cos{x} \cos{hy}-\sin{x} \sin{hy} \)
\(u=\cos⁡{x}\cos{⁡hy} …(1) \)
\(v=-\sin{x} \sin{⁡hy} …(2) \)
Given: x=π
Substituting x=π in (1)
\(v=\cos{⁡π}\cos{⁡hy} \)
\(v= -\cos{⁡hy} \)
We know that cos⁡(hy) ≥ 1
-cos⁡(hy) ≤ -1
v ≤ -1

12. Which of the following is the image in w plane when x=π/2 under the transformation w=cos⁡z?
a) u=0
b) u≥1
c) u≤-1
d) u≥-1
View Answer

Answer: a
Explanation:
Given: w=cos⁡z
\(u+iv=\cos{⁡(x+iy)} \)
\(u+iv=\cos{x}\cos{hy}-\sin{x}\sin{hy} \)
\(u=\cos⁡{x}cos{⁡hy} …(1) \)
\(v=-\sin{x} \sin{⁡hy} …(2) \)
When x= π/2
\(u=\cos{\frac{π}{2}} \cos{hy} \)
u=0

13. Which of the following is the image in w plane when x=-π/2 under the transformation w=cos⁡z?
a) u=0
b) u=1
c) u>1
d) u<1
View Answer

Answer: a
Explanation:
Given: w=cos⁡z
\(u+iv=\cos⁡{(x+iy)} \)
\(u+iv=\cos{x} \cos{hy}-\sin{x} \sin{hy} \)
\(u=\cos⁡{x} \cos{⁡hy} …(1) \)
\(v=-\sin{x} \sin{⁡hy} …(2) \)
When x= -π/2
\(u=\cos⁡{(-\frac{π}{2})} \cos{hy} \)
u=0

14. Which of the following is the image in w plane when x=-3π/2 under the transformation w=cos⁡z?
a) u=0
b) u=1
c) u>>1
d) u<1
View Answer

Answer: a
Explanation:
Given: w=cos⁡z
\(u+iv=\cos⁡{(x+iy)} \)
\(u+iv=\cos{x} \cos{hy}-\sin{x} \sin{hy} \)
\(u=\cos⁡{x}\cos{⁡hy} …(1) \)
\(v=-\sin{x} \sin{⁡hy} …(2) \)
When x= -3π/2
\(u=\cos⁡{(-\frac{3π}{2})} \cos{hy} \)
u=0

15. Which of the following is the image in w plane when x=3π/2 under the transformation w=sin⁡z?
a) u=0
b) u=1
c) u>1
d) u≤-1
View Answer

Answer: d
Explanation:
Given: w=sin⁡z
\(u+iv=\sin{⁡(x+iy)} \)
\(u+iv=\sin{x} \cos⁡{hy}+i\cos{x} \sin⁡{hy} \)
\(u=\sin⁡{x} \cos⁡{hy} …(1) \)
\(v=\cos⁡{x}\sin{⁡hy} …(2) \)
When x= 3π/2
\(u= -\cos⁡{hy} \)
We know that cos⁡(hy)≥1
-cos⁡(hy) ≤ -1
u ≤ -1

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