# Complex Mapping Questions and Answers – Transformation w = sin z

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) $$\frac{u^2}{cos⁡h^2c}+ \frac{v^2}{sin⁡h^2c}=1$$

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) $$\frac{u^2}{cos⁡h^2 c}+ \frac{v^2}{sin⁡h^2 c}=1$$

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) $$\frac{u^2}{cos⁡h^2 c}+ \frac{v^2}{sin⁡h^2 c}=1$$

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) $$\frac{u^2}{cos⁡h^2 c}+ \frac{v^2}{sin⁡h^2 c}=1$$

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) $$\frac{u^2}{cos⁡h^2 c}+ \frac{v^2}{sin⁡h^2 c}=1$$

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) $$\frac{u^2}{cos⁡h^2 c}+ \frac{v^2}{sin⁡h^2 c}=1$$

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

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

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

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

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

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

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

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

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

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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