# Complex Mapping Questions and Answers – Bilinear Transformation

This set of Complex Mapping Multiple Choice Questions & Answers (MCQs) focuses on “Bilinear Transformation”.

1. Which of the following is the bilinear transformation that maps the points z=0, -1, i onto the points w= 0, -1, i respectively?
a) $$\frac{z+1}{z-i}$$
b) $$\frac{1}{z}$$
c) $$\frac{z+1}{z}$$
d) $$\frac{1}{z+1}$$

Explanation:
Given: $$z_1=0, z_2=-1, z_3=i$$
$$w_1=i, w_2=0, w_3=∞$$
Let the required transformation be
$$\frac{(w-w_1 )(w_2-w_3)}{(w-w_3 )(w_2-w_1)} = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$
$$\frac{(w-w_1)}{(w_2-w_1)} = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$
$$\frac{w-i}{0-i}= \frac{(z-0)(-1-i)}{(z-i)(-1-0)}$$
$$\frac{w-i}{-i} = \frac{z}{z-i}(1+i)$$
$$w-i = \frac{z}{(z-i)}(-i+1)$$
$$w= \frac{z}{z-i}(-i+1)+i= \frac{-iz+z+iz+1}{(z-i)} = \frac{z+1}{z-i}$$

2. Which of the following is the bilinear transformation that maps the points z=∞, i, 0 onto the points w=0, i, ∞ respectively?
a) $$\frac{-1}{z}$$
b) $$\frac{1}{z}$$
c) 1
d) -1

Explanation:
Given: $$z_1=∞,z_2= i,z_3=0$$
$$w_1=0,w_2=i,w_3=∞$$
Let the required transformation be
$$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)} = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$
$$\frac{(w-w_1)}{(w_2-w_1)}= \frac{(z_2-z_3)}{(z-z_3)}$$
$$\frac{w-0}{i-0}= \frac{i-0}{z-0}$$
$$w= \frac{-1}{z}$$

3. Which of the following is the bilinear transformation that maps the points z=1, i, -1 onto the points w=0, 1, ∞ respectively?
a) $$\frac{(-i)z+i}{(1)z+1}$$
b) 1
c) i
d) -1

Explanation:
Given:$$z_1=1, z_2= i, z_3=-1$$
$$w_1=0, w_2=1, w_3=∞$$
Let the required transformation be
$$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)} = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$
$$\frac{(w-w_1)}{(w_2-w_1)} = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$
$$\frac{w-0}{1-0} = \frac{(z-1)(i+1)}{(z+1)(i-1)}$$
$$w= \frac{(z-1)(i+1)}{(z+1)(i-1)}$$
$$w= \frac{z-1}{z+1}[-i]$$
$$w= \frac{(-i)z+i}{(1)z+1}$$

4. Which of the following is the bilinear transformation that maps the points z=-2i, i, ∞ onto the points w=0,-3, 1/3 respectively?
a) $$\frac{z+2i}{3z-4i}$$
b) $$\frac{(-i)z+i}{(1)z+1}$$
c) $$\frac{-1}{z}$$
d) 1

Explanation:
Given: $$z_1=1, z_2= i, z_3=-1$$
$$w_1=0, w_2=1, w_3=∞$$
Let the required transformation be
$$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)} = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$
$$\frac{(w-w_1)(w_2-w_3)}{(w_2-w_1)(w_2-w_1)}= \frac{(z-z_1)}{(z_2-z_1)}$$
$$\frac{(w)(\frac{-10}{3})}{(\frac{3w-1}{3})(-3)} = \frac{z+2i}{i+2i}= \frac{z+2i}{3i}$$
$$\frac{10w}{(3w-1)} = \frac{z+2i}{i}$$
10wi = 3wz + 6wi-z-2i
10wi-6wi = 3wz-z-2i
w[4i-3z]= – [z+2i]
$$w= \frac{-[z+2i]}{[4i-3z]} = \frac{z+2i}{3z-4i}$$

5. Which of the following images is represented in z plane for the transformation $$\frac{az+b}{cz+d}$$ if |a|=|c| to transform the unit circle inw plane?
a) $$x[\frac{2b}{a}-\frac{2d}{c}]+ [\frac{b^2}{a^2} -\frac{d^2}{c^2}]=0$$
b) $$x[\frac{2a}{b}-\frac{2d}{c}]+ [\frac{b^2}{a^2}-\frac{d^2}{c^2}]=0$$
c) $$x[\frac{2b}{a}-\frac{2c}{d}]+ [\frac{b^2}{a^2} -\frac{d^2}{c^2}]=0$$
d) $$x[\frac{2b}{a}-\frac{2d}{c}]+ [\frac{b^2}{a^2} +\frac{d^2}{c^2}]=0$$

Explanation:
The transformation $$w= \frac{az+b}{cz+d} = \frac{a(z+\frac{b}{a})}{c(z+\frac{d}{c})}$$
Given: |w|=1(unit circle in w-plane)
$$|\frac{a}{c}||\frac{(z+\frac{b}{a})}{(z+\frac{d}{c})}|=1$$
$$|\frac{(z+\frac{b}{a})}{(z+\frac{d}{c})}|=|\frac{c}{a}|$$
If |a|=|c|, then |a/c|= 1
$$|\frac{(z+\frac{b}{a})}{(z+\frac{d}{c})}|=1 →|z+\frac{b}{a}|=|z+\frac{d}{c}|$$
$$|x+\frac{b}{a}+iy|=|x+\frac{d}{c}+iy|$$
$$\sqrt{(x+\frac{b}{a})^2+y^2)}= \sqrt{(x+\frac{d}{c})^2+y^2)}$$
Squaring, we get
$$(x+\frac{b}{a})^2+y^2=(x+\frac{d}{c})^2+y^2$$
$$(x+\frac{b}{a})^2= (x+\frac{d}{c})^2$$
$$2x\frac{b}{a}+\frac{b^2}{a^2} -2x\frac{d}{c}-\frac{d^2}{c^2} =0$$
$$x[\frac{2b}{a}-\frac{2d}{c}]+ [\frac{b^2}{a^2}-\frac{d^2}{c^2}]=0$$
which represents a straight line in z plane.
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6. Which of the following is the bilinear transformation that maps the points z=-1, 0, 1 onto the points w=0, i, 3i respectively?
a) $$\frac{(-3i)z+(-3i)}{(1)z+(-3)}$$
b) $$\frac{z+(-3i)}{(1)z+(-3)}$$
c) $$\frac{(-3i)z+(i)}{(1)z+(-3)}$$
d) $$\frac{(-3i)z+(-3i)}{(1)z+(-2)}$$

Explanation:
Given: $$z_1=-1,z_2= 0,z_3=1$$
$$w_1=0,w_2=i,w_3=3i$$
Let the required transformation be
$$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}= \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$
$$\frac{(w-0)(i-3i)}{(w-3i)(i-0)}=\frac{[z-(-1)][0-1]}{(z-1)[0-(-1)]}$$
$$\frac{w(-2i)}{(w-3i)(i)} = \frac{(z+1)(-1)}{(z-1)(1)}$$
$$\frac{-2w}{w-3i}=-\frac{z+1}{z-1}$$
$$\frac{2w}{w-3i}=\frac{z+1}{z-1}$$
2wz-2w=wz+w-3zi-3i
w[2z-2-z-1]= -3i(z+1)
w[z-3]=-3i(z+1)
$$w=-3i \frac{(z+1)}{(z-3)}$$
$$w=\frac{(-3i)z+(-3i)}{(1)z+(-3)}$$

7. Which of the following is the bilinear transformation that maps the points z=-2, 0, 2 onto the points w=0, i, -i respectively?
a) $$\frac{(-i)z+(-2i)}{3z+(-2)}$$
b) $$\frac{z+(-2i)}{3z+(-2)}$$
c) $$\frac{(-i)z+(-i)}{3z+(-2)}$$
d) $$\frac{(-i)z+(2i)}{3z+(-2)}$$

Explanation:
Given: $$z_1=-2, z_2= 0, z_3=2$$
$$w_1=0, w_2=i, w_3=-i$$
Let the required transformation be
$$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}= \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$
Let $$A=\frac{w_2-w_3}{w_2-w_1}= \frac{1+i}{i-0}=\frac{2i}{i}=2$$
$$B=\frac{z_2-z_3}{z_2-z_1}= \frac{0-2}{0+2}=-1$$
$$a=Aw_1-Bw_3=(2)(0)-(-1)(-i)= -i$$
$$b=Bw_3 z_1-Aw_1 z_3=(-1)(-i)(-2)-(2)(0)(2)=-2i$$
$$c=A-B=2—1=3$$
$$d=Bz_1-Az_3=(-1)(-2)-(2)(2)= -2$$
We know that $$w= \frac{az+b}{cz+d}, ad-bc≠0$$
$$w= \frac{(-i)z+(-2i)}{3z+(-2)}$$

8. Which of the following is the bilinear transformation that maps the points z=1, i, -1 onto the points w=i,0,-i respectively?
a) $$\frac{iz+1}{(-i)z+1}$$
b) $$\frac{iz}{(-i)z+1}$$
c) $$\frac{iz+1}{(-i)z}$$
d) $$\frac{iz+1}{z+1}$$

Explanation:
Given: $$z_1=-2,z_2= 0,z_3=2$$
$$w_1=0,w_2=i,w_3=-i$$
Let the required transformation be
$$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}= \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$
Let $$A= \frac{w_2-w_3}{w_2-w_1}= \frac{0+i}{0-i}= -1$$
$$B= \frac{z_2-z_3}{z_2-z_1}= \frac{i+1}{i-1}= -i$$
$$a=Aw_1-Bw_3=(-1)(i)-(-i)(-i)= -i+1$$
$$b=Bw_3 z_1-Aw_1 z_3=(-i)(-i)(1)-(-1)(i)(-1)= -1-i$$
$$c=A-B=(-1)—i= -1+i$$
$$d=Bz_1-Az_3=(-i)(1)-(-1)(-1)=-i-1$$
We know that $$w= \frac{az+b}{cz+d}, ad-bc≠0$$
$$w= \frac{(-i+1)z+(-1-i)}{(-1+i)z+(-i-1)}=\frac{iz+1}{(-i)z+1}$$

9. Which of the following is the bilinear transformation which maps z=0 onto w=-i and has 1 and 1 as the invariant points ?
a) $$[\frac{u+iv+i}{(1-v)+iu}][\frac{1-v-iu}{(1-v)^2+u^2}]$$
b) $$[\frac{u}{(1-v)+iu}][\frac{v}{(1-v)^2+u^2}]$$
c) $$[\frac{u+iv+i}{(1-v)+iu}]$$
d) $$[\frac{1-v-iu}{(1-v)^2+u^2}]$$

Explanation:
Given: $$z_1=0,z_2= -1,z_3=1$$
$$w_1=-i,w_2=-1,w_3=1$$
Let the required transformation be
$$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}= \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$
$$Let A=\frac{w_2-w_3}{w_2-w_1}= \frac{-1-i}{-1+i}= \frac{-2}{-1+i}=1+i$$
$$B= \frac{z_2-z_3}{z_2-z_1}=\frac{-1-1}{-1-0}=2$$
$$a=Aw_1-Bw_3=(1+i)(-i)-2(1)=(1+i)(-i)-2(1)=-i+1-2=-i-2$$
$$b=Bw_3 z_1-Aw_1 z_3=(2)(1)(0)-(1+i)(-i)(1)=i-1$$
$$c=A-B=(1+i)-2=i-1$$
$$d=Bz_1-Az_3=(2)(0)-(1+i)(1)= -(1+i)$$
We know that $$w=\frac{az+b}{cz+d}$$
$$w= \frac{(-i-1)z+(i-1)}{(i-1)z+(-1-i)}= \frac{z+(-i)}{(-i)z+1}$$
We know that $$z= \frac{-dw+b}{cw-a}= \frac{-w-i}{-iw-1}=\frac{w+ai}{1+wi}$$
$$z= \frac{u+iv+i}{1+(u+iv)i}$$
$$= \frac{u+iv+i}{1+iu-v}= \frac{u+iv+i}{(1-v)-iu}$$
$$= [\frac{u+iv+i}{(1-v)+iu}][\frac{1-v-iu}{(1-v)^2+u^2}]$$

10. Which of the following is the bilinear transformation that maps the points z=1+i, -i, 2 -1 onto the points w=0, 1, i respectively?
a) $$\frac{(-2i)z+(2i-2)}{(1+i)z+(3i-5)}$$
b) $$\frac{z+(2i-2)}{(1+i)z+(3i-5)}$$
c) $$\frac{(-2i)z+2i}{(1+i)z+(3i-5)}$$
d) $$\frac{z+2}{(1+i)z+(3i-5)}$$

Explanation:
Given: $$z_1=1+i,z_2= -i,z_3=2-i$$
$$w_1=0,w_2=1,w_3=i$$
Let the required transformation be
$$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}= \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$
$$Let A=\frac{w_2-w_3}{w_2-w_1}= \frac{1-i}{1-0}=1-i= \frac{1-i}{1+2i}(1+2i)=\frac{3+i}{1+2i}$$
$$B= \frac{z_2-z_3}{z_2-z_1} = \frac{-i-2+i}{-i-1-i}= \frac{-2}{-1-2i}= \frac{2}{1+2i}$$
$$a=Aw_1-Bw_3=(\frac{3+i}{1+2i})(0)-(\frac{2}{1+2i})(i)= \frac{-2i}{1+2i}$$
$$b=Bw_3 z_1-Aw_1 z_3=(\frac{2}{1+2i})(i)(1+i)-0=\frac{-2+2i}{1+2i}$$
$$c=A-B= \frac{3+i}{1+2i}-\frac{2}{1+2i}= \frac{1+i}{1+2i}$$
$$d=Bz_1-Az_3=(\frac{2}{1+2i})(1+i)-(\frac{3+i}{1+2i})(2-i)=\frac{-5+3i}{1+2i}$$
We know that $$w= \frac{az+b}{cz+d}, ad-bc≠0$$
$$w=\frac{(\frac{-2i}{1+2i})z+(\frac{2i-2}{1+2i})}{(\frac{1+i}{1+2i})z+(\frac{3i-5}{1+2i})}$$
$$w=\frac{(-2i)z+(2i-2)}{(1+i)z+(3i-5)}$$

11. Which of the following is the fixed point of $$w=\frac{2zi+5}{z-4i}$$?
a) i, 2i
b) i, 3i
c) i, 4i
d) i, 5i

Explanation:
The fixed points are given by
$$z=\frac{2zi+5}{z-4i}$$ [Replace w by z]
z2-4iz=2zi+5; z2-6zi-5=0
$$z= \frac{6i± \sqrt{-36+20} }{2}=5i,i$$

12. Which of the following is the invariant point of the bilinear transformation $$\frac{z-1}{z+1}$$?
a) i,-i
b) 1,-1
c) 2i
d) -2i

Explanation:
The invariant points are given by
$$z=\frac{z-1}{z+1}$$
$$z^2+z=z-1$$
$$z^2=-1$$
$$z=±\sqrt{-1}=i,-i$$

13. Which of the following is the invariant point of the bilinear transformation $$w=\frac{1+z}{1-z}$$?
a) ±i
b) ±∞
c) ±1
d) ±2

Explanation:
The invariant points are given by
$$z=\frac{1+z}{1-z}$$
z-z2=1+z
z2=-1
z=±i

14. Which of the following is the invariant point of the bilinear transformation $$w=2-\frac{2}{z}$$?
a) 1±i
b) 2±i
c) 3±i
d) i±4

Explanation:
The invariant points are given by
$$z=2-\frac{2}{z}; z=\frac{2z-2}{z}$$
z2= 2z-2; z2-2z+2=0
$$z=\frac{2±\sqrt{4-8}}{2}= \frac{2±\sqrt{-4}}{2}=\frac{2±2i}{2}=1±i$$

15. Which of the following is the invariant point of the transformation $$w=\frac{3z-4}{z-1}$$?
a) 2,2
b) 1,1
c) 0
d) 3,3

Explanation:
The fixed points are given by
$$z=\frac{3z-4}{z-1}$$
z2-z=3z-4
z2-4z+4=0
(z-2)2=0
z=2,2

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