Probability and complex function: Unit III: Analytic functions

Exercise 3.5 (Bilinear transformation)

Problems with Answer | Analytic functions

Probability and complex function: Unit III: Analytic functions : Exercise 3.5

EXERCISE 3.5

Bilinear transformation

 

1. Find the fixed points of the following mappings

(i) w = z – 1 – i/z+2 Ans. z = -1 ± i√3 + 4i/2

ii) w = 2z -5/z + 4 Ans. z = -1 ± 2i

(iii) w = z -2/z + 3 Ans. z = -1± i

(iv)w = 1/z – 2i Ans. z = i

 (v) w = 5z + 4/2 + 5 Ans. z = ± 2

(vi) w = z2 Ans. z = 0, 1 [ A.U M/J 2014]

 

2. Define bilinear transformation.

 

3. Find the most general bilinear transformation that maps the upper half of the z-plane onto the interior of the unit circle in the w-plane.

 

4. Find the bilinear transformation which maps the points z into w Answers

(1) 2, i, -2; 1, i, −1 w= 3z + 2i/ iz + 6

(2) -i, 0, i ; -1, i, 1 [AU N/D 2015 R-13] w = -i (z – 1/z + 1 )

(3) 0, -i, 2i; 5i, ∞ , 1/3 w = 3z - 5i/iz - 1

(4) 1, -1, ∞ ; 1+i, 1-i, 1 w = z + i/z

(5) 0, 1, ∞ ; i, 1, -i [A.U M/J 2012, 2013] [A.U M/J 2005, 2005][A.U A/M 2008] A.U A/M 2019 R13]  w = z + i/1 + zi

(6) 1, i, -1; 2, i, -2 [AU N/D 2004, 2005] [AU M/J 2016 R-13] w = -(6z – 2i/iz – 3)

(7) 0, 1, ∞ ; -5, -1, 3 [AU D15/J16 R-08] w = 3z - 5/z + 1

(8) i, -1, 1; 0, 1, ∞ w =2z - 2i /(1 + i) (z - 1)

(9) ∞, i, 0 ; 0, -i, ∞ w =  1/z

(10) 0, -i, -1; i, 1, 0  [AU M/J 2008, 2009] w = i (1 + z)/1 - z

 (11) -i, 0, i; ∞, -1, 0 w = z – 1/z + 1

 (12) 0, 1, ∞ ; i, -1; -i w = -i( z + i /z – i)

(13) -1,0,1; -1, -2, 1 [AU N/D 2016 R-08] [AU A/M 2017 R-13] w = z – i / 1- iz

(14) 0, 1, -1; -1,0, ∞ [AU N/D 2016 R-13] w = z -1/z + 1

(15)-2, 0,2; 0, i, -1 [A.U May 2002] w = (-i)z + (-2i)/3z + (-2)

(16) 1, -1, ∞ ; -1, -i, i [AU N/D 2019 (R17)] w = z + (1 + 2i)/(-i)z + (-2-1)

 

Probability and complex function: Unit III: Analytic functions : Tag: : Problems with Answer | Analytic functions - Exercise 3.5 (Bilinear transformation)