Thread: Quick question regarding complex exponentials

1. Quick question regarding complex exponentials

Just wondering, how does $\displaystyle e^{-z} - e^{z} = 1 - e^{iz}$

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

Thanks.

2. Originally Posted by Glitch Just wondering, how does $\displaystyle e^{-z} - e^{z} = 1 - e^{iz}$
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 $\displaystyle z=1$ the RHS is real but the LHS is not.

3. It came from this:

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

I'm trying to simplify it.

4. Originally Posted by Glitch It came from this:
$\displaystyle \frac{(e^{-z} - e^{z})}{i(e^{-z} + e^{z})} = 2i$
I'm trying to simplify it.
What is the goal here?
Are you trying to solve for z ?

5. Well the question this involves is:
Find all solutions $\displaystyle z \in C$ for $\displaystyle tan(iz) = 2i$

So I used the exponential form of sin and cos to try and get an expression for tan.

6. Originally Posted by Glitch Well the question this involves is:
Find all solutions $\displaystyle z \in C$ for $\displaystyle tan(iz) = 2i$
This should be a lesson to you. Had you posted the basic question first, you would have had this reply sooner. That reduces to

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

Can you solve that?

7. 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

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

Can you solve that?
I thought the index laws didn't work for complex exponentials? :/

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