# Thread:

1. ## Exponential Inverse

Ok. I have to show that

f(z)=exp(1-2iz)

does not have an inverse function.

Now I am trying to prove that the f(z) is not one one, but am having a hard time trying to come up with two values, that when inserted in f(z) result in the same answer.

Any help would be great

Thanx

Bex

2. Setting…

$s=f(z)=e^{1-2\cdot i\cdot z}$

… we derive…

$\ln s= 1-2\cdot i\cdot z \rightarrow z=\frac {1-\ln s}{2\cdot i}$

… which is the requested inverse function…

Kind regards

$\chi$ $\sigma$

3. Thanks but you have shown the inverse of f(z), however the question asks me to prove that f(z) DOES NOT have and inverse.

Originally Posted by chisigma
Setting…

$s=f(z)=e^{1-2\cdot i\cdot z}$

… we derive…

$\ln s= 1-2\cdot i\cdot z \rightarrow z=\frac {1-\ln s}{2\cdot i}$

… which is the requested inverse function…

Kind regards

$\chi$ $\sigma$

4. Originally Posted by bex23
Ok. I have to show that

f(z)=exp(1-2iz)

does not have an inverse function.

Now I am trying to prove that the f(z) is not one one, but am having a hard time trying to come up with two values, that when inserted in f(z) result in the same answer.

Any help would be great

Thanx

Bex
$f(z) = f(z + n \pi)$ where n is an integer.

5. Originally Posted by mr fantastic
$f(z) = f(z + n \pi)$ where n is an integer.
So would this give

f(z+n \pi)=exp(1-2iz-2nipi)

And because of the different values of n, f(z) would never be one one and thus have no inverse.
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