Coursera: Analysis of a Complex Kind. 4.4 Mobius Transformations II

Jul 27, 2016

Slide 2 summary of Mobius transformation

Plus final result from previous lesson 4.3,

The particular Mobius transformation f(z) = \frac{z - z_1}{z - z_3} . \frac{z_2 - z_3}{z_2 - z_1}

maps z_1, z_2, z_3 \textrm{ to } 0, 1, \infty

Slide 3 Properties

Composition two Mobius xform is a Mobius xform

Inverse mobius is a Mobius xform

<< proof >>

Slide 4, 5, 6 Deducing mobius transformation that maps a,b,c to d,e,f via 0,1,infinity

Direct Mobius transformation

determine f(z) = \frac{az+b}{cz+d}

when

f:0 → -1

f:i → 0

f:∞ → 1

let a = 1

f(z) = \frac{z+b}{cz+d}

Consider mapping to zero. f(i) = 0 therefore numerator = 0

i + b = 0, b = – i

f(z) = \frac{z-i}{cz+d}

Consider mapping from infinity. f(∞) = 1

f(z) = \frac{z-i}{cz+d}. As z increases then z dominates over constant and function tends to f(z) = \frac{z}{cz} and since f(∞) = 1, then c = 1.

f(z) = \frac{z-i}{z+d}. Consider final mapping f(0) = -1. Substitute into latest expression.

f(z) = \frac{0-i}{0+d} = -1 = \frac{-i}{+d}, therefore d = i.

Finally, f(z) = \frac{z-i}{z+i}

Slide 7

z -> az   (rotation and dilation)

z -> z + b (translation)

z -> 1/z   Inversion

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