# Complex Analysis Questions and Answers – Mapping w = z^2

This set of Complex Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Mapping w = z^2”.

1. Which of the following is the image of the hyperbola x2-y2=10 under the transformation w= z2 if w = u + i v?
a) u=10
b) u=15
c) u=30
d) u=20

Explanation:
w= z2
u+iv= (x+iy)2
= x2– y2+ i2xy
i.e. u= x2-y2…………(1)
v=2xy…………(2)
Given: x2-y2=10
u=10
Hence, the image of the hyperbola x2– y2=10 in the z plane is mapped into u=10 in the w plane, which is a straight line.

2. Which of the following regions is the region of w plane into which the circle |z-1|= 1 is mapped by the transformation w= z2 ?
a) R=2[1+sin⁡∅]
b) R=[1+sin⁡∅]
c) R=[1+cos⁡∅]
d) R=2[1+cos⁡∅]

Explanation:
In polar form,z =r e, w = R e
Given: |z-1|= 1
i.e.|re-1|=1
|rcos⁡θ+ irsin⁡θ-1|=1
|(rcos⁡θ-1)+ irsin⁡θ|=1
(rcos⁡θ-1)2+ (rsin⁡θ)2= 12
r2cos2θ+ 1- 2rcos⁡θ+ r2sin2 θ=1
r2 [cos2θ+ sin2θ]= 2rcos⁡θ
r2=2rcos⁡θ
r=2 cos⁡θ…………(1)
Given: w= z2
Re(i∅)= (re)2
Rei∅=r2 ei2θ
R= r2, ∅=2θ…………(2)
(1)→ r2= (2cos⁡θ)2
r2=4cos2θ
=4 $$[\frac{1+cos⁡2θ}{2}]$$
r2=2[1+cos⁡2θ]
R=2 [1+cos∅] (From (2))

3. Which of the following images is an image under the mapping w=z2 of the triangular region bounded by y=1, x=1 and x+y= 1?
a) $$u^2= -2(v-\frac{1}{2})$$
b) $$u^2=2(v- \frac{1}{2})$$
c) $$u^2=(v- \frac{1}{2})$$
d) $$u^2=(v+ \frac{1}{2})$$

Explanation:
In Z plane, given lines are x=1, y=1, x+y=1
Given: w= z2
u+iv= (x+iy)2
u+iv= x2– y2+ 2xyi
Equating the real and imaginary parts, we get
u= x2– y2…………(1)
v=2 xy…………(2)

When x=1 When y=1
(1)→u=1- y2…(3)
(2)→v=2y…(4)
(1)→u= x2-1…(5)
(2)→v=2x…(6)
(4)→v2=4y2
v2=4(1-u) by (3)
i.e. v2=-4(u-1)
(6)→ v2= 4x2
=4(u+1) by(5)

When x + y =1
(1) →u= (x+y)(x-y)
u=x-y [x+y=1] $$u= \sqrt{(x+y)^2-4xy}$$ (Identity)
$$u= \sqrt{1-2v}$$
$$u^2=1-2v = -2 (v-\frac{1}{2})$$
The image of x =1 is v2= -4(u-1)
The image of y =1 is v2=4(u+1)
The image of x+y=1 is $$u^2= -2(v-\frac{1}{2})$$

4. Which of the following images is an image of |z-2i| = 2 under the transformation $$w= \frac{1}{z}$$?
a) $$v= \frac{-1}{2}$$
b) $$v= \frac{1}{2}$$
c) $$v= \frac{-1}{4}$$
d) $$v= \frac{1}{4}$$

Explanation:
Given:|z-2i|= 2……(1) is a circle
Centre = 2i i.e., (0,2)
Given: $$w=\frac{1}{z}→z= \frac{1}{w}$$

$$(1)→|\frac{1}{w}-2i|= 2$$
|1-2wi|= 2|w|
|1-2(u+iv)i|= 2 |u+iv|
|1-2ui+2v|= 2|u+vi|
|1+2v-2ui|= 2 |u+iv|
$$\sqrt{(1+2v)^2+(-2u)^2}=2 \sqrt{u^2+v^2}$$
$$(1+2v)^2+ 4u^2= 4(u^2+ v^2)$$
1+4v=0
$$v=\frac{-1}{4}$$, which is a straight line in w-plane.
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5. Which of the following images is the image of the circle |z-1|=1 in the complex plane under the mapping $$w= \frac{1}{z}?$$
a) $$u=\frac{-1}{2}$$
b) $$u= \frac{1}{2}$$
c) $$u= \frac{-1}{4}$$
d) $$u= \frac{1}{4}$$

Explanation:
Given: |z-1|= 1……(1) is a circle
Centre=1, i.e.,(1,0)
Given: $$w= \frac{1}{z}→z= \frac{1}{w}$$

$$(1)→|\frac{1}{w}-1|=1$$
|1-w|=|w|
|1-(u+iv)|=|u+iv|
|1-u-iv|=|u+iv|
$$\sqrt{√(1-u)^2+(-v)^2} = \sqrt{u^2+v^2}$$
$$(1-u)^2+ v^2 = u^2 + v^2$$
$$1+u^2- 2u+ v^2= u^2+ v^2$$
2u=1
$$u=\frac{1}{2},$$ which is a straight line in the w-plane.

6. Which of the following images is an image of the infinite strips 1/4<y<1/2 under the transformation w = 1/z?
a) Region between circle of centre(0,-2)and radius 2 and circle of centre (0,-1) and unit radius
b) Region between circle of centre(0,2) and radius 2 and circle of centre (0,1) and unit radius
c) Region between circle of centre(0,1) and radius 1 and circle of centre (0,-1)and unit radius
d) Region between circle of centre(1,0) and radius 1 and the circle of centre (2,0) and radius 2

Explanation:
$$w= \frac{1}{z} (Given)$$
i.e., $$z= \frac{1}{w}$$
$$z= \frac{1}{u+iv}= \frac{u-iv}{(u+iv)(u-iv)}= \frac{u-iv}{u^2+v^2}$$
$$x+iy= \frac{u-iv}{u^2+v^2}= [\frac{u}{u^2+v^2}] + i [\frac{-v}{u^2+v^2}]$$
i.e., $$x= \frac{u}{u^2+v^2}…(1), y=\frac{-v}{u^2+v^2}…(2)$$
Given strip is 1/4 $$u^2+v^2+4v=0$$
$$u^2+(v+2)^2-4=0$$
$$u^2+(v+2)^2=4$$ …(3)
It is a circle whose centre is (0,-2)in the w-plane and radius is 2
When y= 1/2
$$\frac{1}{2}= \frac{-v}{u^2+v^2} by (2)$$
$$u^2+v^2= -2v$$
$$u^2+v^2+2v=0$$
$$u^2+ (v+1)^2-1=0$$
$$u^2+(v+1)^2=1$$ …(4)
It is a circle whose centre is (0,-1)in the w-plane and unit radius

Hence the infinite strip 1/42+(v+1)2=1 and u2+(v+2)2=4 in the w plane.

7. Which of the following images is an image of the infinite strips 0 <y< 1/2 under the transformation w=1/z?
a) Region outside the circle u2+(v+1)2=1
b) Region outside the circle (u-1)2+v2=1
c) Region outside the circle u2+v2=1
d) Region outside the circle (u-1)2+v2=1

Explanation:
Given: $$w= \frac{1}{z}$$
i.e., $$z= \frac{1}{w}$$
$$z= \frac{1}{u+iv}= \frac{u-iv}{(u+iv)(u-iv)}=\frac{u-iv}{u^2 + v^2}$$
$$x+iy=(\frac{u}{u^2+v^2})+ i(\frac{-v}{u^2+ v^2})$$
i.e., $$x=(\frac{u}{u^2+ v^2})…(1), y=(\frac{-v}{u^2+v^2})…(2)$$
Given strip is 0 <y< 1/2

when y=0
→v=0 by (2)
When $$y= \frac{1}{2},$$ we get u2+(v+1)2=1 by (4)
Hence, the infinite strip 0 2+(v+1)2=1 in the lower half of the w-plane.

8. Which of the following is the image of x =2 under the transformation $$w= \frac{1}{z}$$?
a) $$(u-\frac{1}{4})^2+v^2=(\frac{1}{4})^2$$
b) $$(u-\frac{1}{4})^2+(v-1)^2=(\frac{1}{4})^2$$
c) $$(u-\frac{1}{3})^2+v^2=(\frac{1}{4})^2$$
d) $$(u-\frac{1}{2})^2+v^2=(\frac{1}{2})^2$$

Explanation:
Given: $$w= \frac{1}{z}$$
i.e., $$z= \frac{1}{w}$$
$$z= \frac{1}{u+iv}= \frac{u-iv}{(u+iv)(u-iv)}= \frac{u-iv}{u^2+ v^2}$$
$$x+iy=(\frac{u}{u^2+v^2})+ i(\frac{-v}{u^2+ v^2})$$
i.e., $$x=(\frac{u}{u^2+ v^2})…(1), y=(\frac{-v}{u^2+v^2})…(2)$$
Given: x=2 in the z-plane
$$2= \frac{u}{u^2+v^2}$$ by (1)
$$2(u^2+v^2 )= u$$
$$u^2+ v^2-\frac{1}{2} u=0$$
$$(u-\frac{1}{4})^2+ v^2-\frac{1}{16}=0$$
$$(u-\frac{1}{4})^2+v^2=(\frac{1}{4})^2$$
It is a circle whose centre is (1/4, 0)and radius 1/4.
x=2 in the z-plane is transformed into a circle $$(u-\frac{1}{4})^2+v^2=(\frac{1}{4})^2$$ in the w-plane.

9. Which of the following is the image of a circle containing the origin in the XY plane under the transformation $$w= \frac{1}{z}$$?
a) A straight line
b) A circle
c) An ellipse
d) A parabola

Explanation:
Given:$$w= \frac{1}{z}$$
i.e., $$z= \frac{1}{w}$$
$$z= \frac{1}{u+iv}= \frac{u-iv}{(u+iv)(u-iv)}= \frac{u-iv}{u^2 + v^2}$$
$$x+iy= (\frac{u}{u^2+v^2}+ i(\frac{-v}{u^2+ v^2})$$
i.e., $$x=(\frac{u}{u^2+ v^2})…(1), y=(\frac{-v}{u^2+v^2})…(2)$$
Given: $$a(x^2+y^2)+bx+cy=0$$
Substituting (1) and (2), we get
$$a[\frac{u^2}{(u^2+v^2)^2} + \frac{v^2}{(u^2+v^2)^2}]+b[\frac{u}{u^2+v^2}]+ c[\frac{-v}{u^2+v^2}]=0$$
$$a[\frac{u^2+v^2}{(u^2+v^2)^2}]+\frac{bu-cv}{u^2+v^2}=0$$
$$a \frac{1}{(u^2+v^2)}+\frac{bu-cv}{u^2+v^2}=0$$
a+ bu- cv= 0
Therefore, the image of circle passing through the origin in the XY plane is a straight line in the w-plane.

10. Which of the following is the image of 1< x < 2 under the mapping w = 1/z?
a) Region between the circles $$(u-\frac{1}{2})^2+v^2=(\frac{1}{2})^2 and (u-\frac{1}{4})^2+v^2=(\frac{1}{4})^2$$
b) Region between the circles $$(u-\frac{1}{2})^2+v^2= 2^2 and (u-\frac{1}{3})^2+v^2= 3^2$$
c) Region between the circles $$(u-\frac{1}{5})^2+v^2= 3^2 and (u-\frac{1}{2})^2+v^2= 4^2$$
d) Region between the circles $$(u-\frac{1}{2})^2+(v-1)^2= (\frac{1}{2})^2 and (u-\frac{1}{4})^2+ (v-2)^2=(\frac{1}{3})^2$$

Explanation:
Given: $$w= \frac{1}{z}$$
i.e., $$z= \frac{1}{w}$$
$$z= \frac{1}{u+iv}= \frac{u-iv}{(u+iv)(u-iv)}=\frac{u-iv}{u^2+v^2}$$
$$x+iy=(\frac{u}{u^2+v^2})+ i (\frac{-v}{u^2+ v^2})$$
i.e., $$x=(\frac{u}{u^2+ v^2})…(1), y=(\frac{-v}{u^2+v^2})…(2)$$
Given: 1 $$u^2+v^2=u$$
$$u^2+v^2-u=0$$
$$(u-\frac{1}{2})^2+v^2-\frac{1}{4}=0$$
$$(u-\frac{1}{2})^2+v^2=(\frac{1}{2})^2…(3)$$
It is a circle whose centre is (1/2,0) and radius is 1/2
When x = 2
$$2= \frac{u}{u^2+v^2}\, by(1)$$
$$2(u^2+v^2 )= u$$
$$u^2+v^2-\frac{u}{2}=0$$
$$(u-\frac{1}{4})^2+v^2-(\frac{1}{4})^2=0$$
$$(u-\frac{1}{4})^2+v^2= (\frac{1}{4})^2…(4)$$
It is a circle whose centre is $$(\frac{1}{4},0)$$ and radius is $$\frac{1}{4}$$
Hence the infinite strip 1<x<2 is transformed into the region in between the circles (3) and (4) in the w-plane.

11. Which of the following is the image of the hyperbola x2-y2=1 under the transformation w=1/z?
a) Cardioids
b) Lemniscates
c) Parabola
d) Hyperbola

Explanation:
Given:$$w=\frac{1}{z} →z=\frac{1}{w}$$
$$x+iy= \frac{1}{Re^{i∅}}$$
$$x+iy = \frac{1}{R}e^{-i∅}= \frac{1}{R}[cos⁡∅-isin⁡∅]$$
$$x=\frac{1}{R}cos⁡∅, y= \frac{-1}{R} sin⁡∅$$
Given: x2-y2=1
$$[\frac{1}{R}cos⁡∅]^2-[\frac{-1}{R}sin⁡∅]^2=1$$
$$\frac{cos^2∅-sin^2∅}{R^2} =1$$
$$cos⁡2∅= R^2 i.e., R^2=cos⁡2∅$$
It is a lemniscate.

12. The transformation $$w=\frac{1}{z}$$ transforms all circles and straight lines in the z-plane into circles or straight lines in the w-plane.
a) True
b) False

Explanation:
Given: $$w=\frac{1}{z}$$
i.e., $$z=\frac{1}{w}$$
Now, w= u+iv
$$z= \frac{1}{w}=\frac{1}{u+iv}= \frac{1}{u+iv}\frac{u-iv}{u-iv}= \frac{u-iv}{u^2+v^2}$$
i.e., $$x+iy= \frac{u}{u^2+v^2}- i\frac{v}{u^2+v^2}$$
$$x= \frac{u}{u^2+ v^2} …(1) y= \frac{-v}{u^2+v^2} …(2)$$
The general equation is
$$a(x^2+y^2)+2gx+2fy+c=0…(3)$$
$$a[\frac{u^2}{(u^2+v^2)^2} + \frac{v^2}{(u^2+v^2)^2}]+ 2g[\frac{u}{(u^2+v^2)}]+2f[\frac{-v}{u^2+v^2}]+c=0$$
The transformed equation is
$$c(u^2+v^2 )+ 2gu-2fv+a=0…(4)$$
(i)a≠0, c≠0→Circles not passing through the origin in z plane map onto circles not passing through the origin in the w plane.
(ii) a≠0,c=0 →Circles through the origin in z plane map onto straight line not through the origin in the w plane.
(iii) a=0,c≠0→The straight lines not through the origin in z plane map onto the circles through the origin in the w plane.
(iv) a=0,c=0 →The straight lines through the origin of z plane onto the straight lines through the origin in the w plane.

13. Which of the following is the image of the infinite series 0 <y< 1/4 under the transformation w=1/z?
a) Region outside the circle (u-1)2+v2=1
b) Region outside the circle u2+v2=1
c) Region outside the circle u2+(v+2)2=4
d) Region outside the circle u2+(v+1)2=1

Explanation:
Given: $$w= \frac{1}{z}$$
i.e., $$z=\frac{1}{w}$$
$$z= \frac{1}{u+iv}= \frac{u-iv}{(u+iv)(u-iv)}=\frac{u-iv}{u^2+v^2}$$
$$x+iy=(\frac{u}{u^2+v^2})+ i(\frac{-v}{u^2+ v^2})$$
i.e., $$x= (\frac{u}{u^2+ v^2})…(1), y=(\frac{-v}{u^2+v^2})…(2)$$
Given strip is 0 $$u^2+v^2= -4v$$
$$u^2+ v^2+4v=0$$
$$u^2+(v+2)^2-4=0$$
$$u^2+(v+2)^2=4 … (3)$$
It is a circle whose centre is at (0,-2) in the w- plane and radius 2.

Hence, the infinite strip 0<y<1/4 is mapped into the region outside the circle u^2+〖(v+2)〗^2=4 in the lower half of the w-plane.

14. Which of the following is the image of x=3 under the transformation w=1/z?
a) $$(u-\frac{1}{4})^2+v^2=(\frac{1}{4})^2$$
b) $$(u-\frac{1}{4})^2+(v-1)^2=(\frac{1}{4})^2$$
c) $$(u-\frac{1}{3})^2+v^2=(\frac{1}{4})^2$$
d) $$(u-\frac{1}{9})^2+v^2= (\frac{1}{9})^2$$

Explanation:
Given:$$w= \frac{1}{z}$$
i.e., $$z=\frac{1}{w}$$
$$z= \frac{1}{u+iv}= \frac{u-iv}{(u+iv)(u-iv)}=\frac{u-iv}{u^2+v^2}$$
$$x+iy=(\frac{u}{u^2+v^2})+ i(\frac{-v}{u^2+ v^2})$$
i.e., $$x=(\frac{u}{u^2+ v^2})…(1), y=(\frac{-v}{u^2+v^2})…(2)$$
Given: x = 3 in the z-plane
$$3= \frac{u}{u^2+v^2} By (1)$$
$$3(u^2+v^2 )= u$$
$$u^2+v^2- \frac{1}{3} u=0$$
$$(u-\frac{1}{9})^2+v^2= (\frac{1}{9})^2$$
It is a circle whose centre is $$(\frac{1}{9}, 0)$$ and radius $$\frac{1}{9}$$
x=3 in the z-plane is transformed into a circle $$(u-\frac{1}{9})^2+v^2= (\frac{1}{9})^2$$ in the w plane.

15. Which of the following is the image of 1<x<3 under the mapping w= 1/z?
a) Region between the circles $$(u-\frac{1}{2})^2+v^2=(\frac{1}{2})^2 and (u-\frac{1}{4})^2+v^2=(\frac{1}{4})^2$$
b) Region between the circles $$(u-\frac{1}{2})^2+v^2= 2^2 and (u-\frac{1}{3})^2+v^2= 3^2$$
c) Region between the circles $$(u-\frac{1}{2})^2+v^2= (\frac{1}{2})^2 and (u-\frac{1}{9})^2+v^2= (\frac{1}{9})^2$$
d) Region between the circles $$(u-\frac{1}{2})^2+v^2= 2^2 and (u-\frac{1}{9})^2+v^2= (\frac{1}{9})^2$$

Explanation:
Given:$$w= \frac{1}{z}$$
i.e., $$z= \frac{1}{w}$$
$$z= \frac{1}{u+iv}= \frac{u-iv}{(u+iv)(u-iv)}=\frac{u-iv}{u^2+v^2}$$
$$x+iy = (\frac{u}{u^2+v^2})+ i(\frac{-v}{u^2+ v^2})$$
i.e., $$x=(\frac{u}{u^2+ v^2})…(1), y=(\frac{-v}{u^2+v^2})…(2)$$
Given:1 <x< 3
when x=1
$$1=\frac{u}{u^2+v^2} by (1)$$
$$u^2+v^2=u$$
$$u^2+v^2-u=0$$
$$(u-\frac{1}{2})^2+v^2= (\frac{1}{2})^2…(3)$$
It is a circle whose centre is (1/2,0) and radius is 1/2
when x=3
$$3=\frac{u}{u^2+v^2} by (1)$$
$$3(u^2+v^2)= u$$
$$u^2+v^2-\frac{u}{3}=0$$
$$(u-\frac{1}{9})^2+v^2= (\frac{1}{9})^2…(4)$$
It is a circle whose centre is (1/9,0) and radius is 1/9
Hence, the infinite strip 1 <x< 3 is transformed into the region in between the circles (3) and (4) in the w plane.

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