Problems based on fixed points or invariant points

 PROBLEMS BASED ON FIXED POINTS OR INVARIANT POINTS

Example 3.5.1. Find the fixed points of w = 2zi+ 5 /  z-4i

Solution: The fixed points are given by

z = 2zi + 5/z-4i-  [Replace w by z]

z² - 4zi = 2zi + 5 ; z² - 6iz - 5  = 0

z = 6i ± √-36 + 20 /2 z = 5i, i

Example 3.5.2 Find the invariant points of the bilinear transformation z – 1/z + 1 [A.U A/M 2015 R13] [A.U A/M 2019 R08]

Solution: The invariant (fixed) points are given by

z = z – 1/z + 1

z² + z = z −1

z² = -1

z  = ± √ -1 = i, - i

Example 3.5.3 Find the invariant point of the bilinear transformation w = 1 + z / 1-z  [A.U. N/D 2007, N/D 2011]

Solution: The invariant points are given by [Put w = z]

z = 1+ z / 1 - z

z - z² = 1 + z

z² = -1

z = ± i

Example 3.5.4. Obtain the invariant points of the transformation w = 2 – 2/z’[Anna, May 1996]

^Solution: The invariant points are given by

z = 2 – 2/ z;  z =  2z – 2 / z

z² = 2z – 2 ;  z² - 2z + 2 = 0 ;

z = 2 ± √4 – 8/2 = 2 ± √-4 /2 = 2 ± 2i /2 = 1 ± i

Example 3.5.5. 

Find the fixed points of (a) w = 3z – 4/z – 1, (b) = 1/z  [A.U. M/J 2006] [A.U N/D 2019 R-17, A.U A/M 2019 R-17]

Solution :

Example 3.5.6.Find the fixed point of the transformation w = 6z – 9/z [A.U. N/D 2005] [A.U M/J 2013 R-08]

Solution: The fixed points are given by replacing w = z

i.e., 6z – 9/z ⇒ z = 6z – 9/z

z² = 6z - 9

z² - 6z +9 = 0

(z - 3)² = 0

z = 3, 3

The fixed points are 3, 3

Example 3.5.7 Find the invariant points of the transformation w = 2z + 6 /z + 7 [A.U M/J 2009]

Solution: The invariant (fixed) points are given by

z = 2z + 6/z + 7

z² + 7z = 2z + 6

z² + 5z - 6=0

Example 3.5.8.Find the invariant points of ƒ (z) = z². [A.U M/J 2014 R-13]

Solution: The invariant points are given by z = w =  f (z)

 (i.e.,) z = z²

⇒ z² - 2 = 0

⇒ z (z − 1) = 0

⇒ z = 0 , z = 1

Example 3.5.9. Find the fixed points under the transformation w = 2z – 5/z + 4’ [A.U. P.T. A/M 2003] [A.U April 2016 R-15 U.D]

Solution: The fixed points are given by

z = 2z -5/2 + 4

z² + 4z = 2z – 5

z² + 2z + 5 = 0

z = -2 ± √4 – 20/2 = -2 ± 4i/2 = -1 ± 2i

z = -1 + 2i , -1 - 2i

Example 4.5.10. Find the invariant points of a function f (z) = z³ + 7z/7 - 6zi

 [A.U D15/J16 R-13]

Solution : Given: w =  f (z) = z³ + 7z/7 - 6zi

The invariant points are given by

z = z³ + 7z /7 - 6zi ⇒ 1 = z² + 7/7 - 6zi

7 - 6zi = z² + 7

-6zi  = z² ⇒ z² + 6zi = 0 ⇒ z (z + 6i) = 0

z = 0, z= -6i