Exponential Inverse

**bex23** · March 20th 2009, 07:27 AM

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

**chisigma** · March 20th 2009, 07:43 AM

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

$\chi$ $\sigma$

**bex23** · March 21st 2009, 05:33 AM

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

$\chi$ $\sigma$

**mr fantastic** · March 30th 2009, 05:12 AM

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.

**bex23** · March 30th 2009, 05:35 AM

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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