Inverse of a complex function

Oct 2012
3
0
London
hi, i have been set the following problem, where z is a complex number:


What is the domain of the function f(z) = (3z+1)/(z+i) ?

Prove that f, defined in the domain of f, has an inverse function (f-1),

i.e. check all necessary properties for the existence of an inverse function.

Determine f-1 and its domain.

I have managed to find the domain and image of the function, and therefore the domain of f-1,

but the problem im having is showing that the function is 1-1 (or injective), in order to show that it has an inverse.

any help would be greatly appreciated!
 
Sep 2012
1,061
434
Washington DC USA
Straight algebra and the definition works:

\(\displaystyle \text{If }z_1, z_2 \in \mathbb{C} \backslash \{-i\}, \text{ and } f(z_1) = f(z_2), \text{ then }\)

\(\displaystyle \frac{3z_1 + 1}{z_1 + i} = \frac{3z_2 + 1}{z_2 + i}, \text{ so }\)

\(\displaystyle (3z_1 + 1)(z_2 + i) = (3z_2 + 1)(z_1 + i), \text{ so }\)

\(\displaystyle 3z_1z_2 + z_2 +3iz_1 + i = 3z_1z_2 + z_1 +3iz_2 + i, \text{ so }\)

\(\displaystyle z_2 +3iz_1 = z_1 +3iz_2, \text{ so }\)

\(\displaystyle z_2-z_1 = 3i(z_2-z_1), \text{ so }\)

\(\displaystyle 0 = (-1+3i)(z_2-z_1), \text{ so }\)

\(\displaystyle 0 = z_2-z_1, \text{ so }\)

\(\displaystyle z_1 = z_2. \text{ Thus f is injective on its domain.}\)
 
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Nov 2012
1
0
Hertfordshire
This message looks very similar to the coursework I set my Complex Analysis class. Paul please come see me in my office when I return. This is a clear violation of the academic code of conduct.