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) \( \frac{1}{z+1} \)
View Answer
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
View Answer
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
View Answer
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
View Answer
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) \( x[\frac{2b}{a}-\frac{2d}{c}]+ [\frac{b^2}{a^2} +\frac{d^2}{c^2}]=0 \)
View Answer
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.
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) \(\frac{(-3i)z+(-3i)}{(1)z+(-2)}\)
View Answer
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) \(\frac{(-i)z+(2i)}{3z+(-2)}\)
View Answer
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) \( \frac{iz+1}{z+1} \)
View Answer
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) \([\frac{1-v-iu}{(1-v)^2+u^2}]\)
View Answer
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) \(\frac{z+2}{(1+i)z+(3i-5)} \)
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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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