Inverse of a complex function

**paulluap1991** · October 29th 2012, 05:07 AM

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!

**johnsomeone** · October 29th 2012, 10:29 AM

Straight algebra and the definition works:

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

$\frac{3z_1 + 1}{z_1 + i} = \frac{3z_2 + 1}{z_2 + i}, \text{ so }$

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

$3z_1z_2 + z_2 +3iz_1 + i = 3z_1z_2 + z_1 +3iz_2 + i, \text{ so }$

$z_2 +3iz_1 = z_1 +3iz_2, \text{ so }$

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

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

$0 = z_2-z_1, \text{ so }$

$z_1 = z_2. \text{ Thus f is injective on its domain.}$

**ccarvalho2** · November 5th 2012, 03:03 PM

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.

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