Inverse of a Complex function

**Len** · September 26th 2011, 09:33 PM

Let X belong to C denote the set of points z such that |z|< 2, and define

f : X --> C by the formula

f(z)= (3z+i)/(3-iz)

Find a formula for the inverse map g=f^-1

Could really use some help, complex numbers are the introduction to my course and I'm kind of stuck. Any and all help is appreciated!

**Prove It** · September 26th 2011, 09:37 PM

Originally Posted by Len

Let X belong to C denote the set of points z such that |z|< 2, and define

f : X --> C by the formula

f(z)= (3z+i)/(3-iz)

Find a formula for the inverse map g=f^-1

Could really use some help, complex numbers are the introduction to my course and I'm kind of stuck. Any and all help is appreciated!

The inverse function will have the f and z values swap...

**Deveno** · September 27th 2011, 12:14 AM

to amplify ProveIt's response, let f(z) = w, so we have

w = (3z+i)/(3-iz).

now solve for z in terms of w. the result will be z = g(w), for some function g, which is the function you are looking for.

**chisigma** · September 27th 2011, 06:56 AM

Originally Posted by Deveno

to amplify ProveIt's response, let f(z) = w, so we have

w = (3z+i)/(3-iz).

now solve for z in terms of w. the result will be z = g(w), for some function g, which is the function you are looking for.

To amplify Deveno's response, let...

$w=f(s)= c_{1}\ s + c_{2}\ s^{2} + c_{3}\ s^{3} +...\ ;\ c_{1}\ne 0$ (1)

... which is analytic for $|s|<r$ , its inverse function is...

$s=f^{-1}(w)= d_{1}\ w + d_{2}\ w^{2}+ d_{3}\ w^{3}+...$ (2)

... where...

$d_{n}= \frac{1}{n!}\ \lim_{s \rightarrow 0} \frac{d^{n-1}}{d s^{n-1}}\ (\frac{s}{f(s)})^{n}$ (3)

If $w(z)= \frac{3 z+i}{3-i z}$ a function like (1) can be obtained setting $s=z+\frac{i}{3}$ ...

$w(s)= \frac{s}{(1-\frac{i}{6})-i\ \frac{s}{3}}$ (4)

Now You have to compute the coefficients $d_{n}$ using (3)...

$\chi$ $\sigma$

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