Quick question regarding complex exponentials

**Glitch** · April 30th 2011, 06:45 AM

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.

**Plato** · April 30th 2011, 07:08 AM

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.

**Glitch** · April 30th 2011, 07:22 AM

It came from this:

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

I'm trying to simplify it.

**Plato** · April 30th 2011, 07:35 AM

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 ?

**Glitch** · April 30th 2011, 07:42 AM

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.

**Plato** · April 30th 2011, 01:10 PM

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?

**Glitch** · April 30th 2011, 07:09 PM

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? :/

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