# Quick question regarding complex exponentials

• Apr 30th 2011, 07:45 AM
Glitch
Quick question regarding complex exponentials
Just wondering, how does $e^{-z} - e^{z} = 1 - e^{iz}$

My tutor did this, but I'm not sure how he did it.

Thanks.
• Apr 30th 2011, 08:08 AM
Plato
Quote:

Originally Posted by Glitch
Just wondering, how does $e^{-z} - e^{z} = 1 - e^{iz}$
My tutor did this, but I'm not sure how he did it.

You need to double cheek your tutor's notes.
As written that is not true.
If $z=1$ the RHS is real but the LHS is not.
• Apr 30th 2011, 08:22 AM
Glitch
It came from this:

$\frac{(e^{-z} - e^{z})}{i(e^{-z} + e^{z})} = 2i$

I'm trying to simplify it.
• Apr 30th 2011, 08:35 AM
Plato
Quote:

Originally Posted by Glitch
It came from this:
$\frac{(e^{-z} - e^{z})}{i(e^{-z} + e^{z})} = 2i$
I'm trying to simplify it.

What is the goal here?
Are you trying to solve for z ?
• Apr 30th 2011, 08:42 AM
Glitch
Well the question this involves is:
Find all solutions $z \in C$ for $tan(iz) = 2i$

So I used the exponential form of sin and cos to try and get an expression for tan.
• Apr 30th 2011, 02:10 PM
Plato
Quote:

Originally Posted by Glitch
Well the question this involves is:
Find all solutions $z \in C$ for $tan(iz) = 2i$

This should be a lesson to you. Had you posted the basic question first, you would have had this reply sooner. That reduces to

$e^{-z}-e^{z}=-2e^{-z}-2e^{z}$
$1-e^{2z}=-2-2e^{2z}$
$3+e^{2z}=0$.

Can you solve that?
• Apr 30th 2011, 08:09 PM
Glitch
Quote:

Originally Posted by Plato
This should be a lesson to you. Had you posted the basic question first, you would have had this reply sooner. That reduces to

$e^{-z}-e^{z}=-2e^{-z}-2e^{z}$
$1-e^{2z}=-2-2e^{2z}$
$3+e^{2z}=0$.

Can you solve that?

I thought the index laws didn't work for complex exponentials? :/