# Math Help - complex exponentials

1. ## complex exponentials

May be its not the right place to ask this, but mods can move this to other parts of the forum(i encountered it while leaning diff equations)...

I would like to know how to find $\Re(-\frac{2}{e^{ix}+i})$ in trigonometric form or polar form, so please try to show all the steps.

It would also be very helpful if someone could post a link to learn converting complex exponentials to trignometric functions, or just how to find out the real part of the complex exponential in trignometric form.

Thanks

2. ## Re: complex exponentials

You need to make use of Euler's formula: \displaystyle \begin{align*} e^{i\,\theta} = \cos{(\theta)} + i\sin{(\theta)} \end{align*}.

3. ## Re: complex exponentials

Originally Posted by Prove It
You need to make use of Euler's formula: \displaystyle \begin{align*} e^{i\,\theta} = \cos{(\theta)} + i\sin{(\theta)} \end{align*}.
I know, but how do i do it???
please solve it fully or post a link where similar problems are solved...

4. ## Re: complex exponentials

Originally Posted by earthboy
I know, but how do i do it???
please solve it fully or post a link where similar problems are solved...
I will do neither. YOU will try this. When you write down the exponential in terms of its real and imaginary parts using Euler's formula, you will have a complex fraction. You should know that you simplify fractions by multiplying top and bottom by the conjugate.

5. ## Re: complex exponentials

Originally Posted by Prove It
I will do neither. YOU will try this. When you write down the exponential in terms of its real and imaginary parts using Euler's formula, you will have a complex fraction. You should know that you simplify fractions by multiplying top and bottom by the conjugate.
$\frac{-2}{e^{ix}+i}$,which by applying euler's formula becomes $\frac{-2}{\cos x+i \sin x +i}=\frac{-2}{\cos x+i(\sin x+1)}$
multiplying it by conjugate becomes $\frac{-2(\cos x-i(\sin x+1))}{(\cos x+i(\sin x+1))(\cos x-i(\sin x+1)}$
as we are dealing with the real part only it becomes $\frac{-2\cos x}{2\sin x +2}=-\frac{\cos x}{\sin x+1}$
right???
Thanks for the hint though....and to euler too..

6. ## Re: complex exponentials

Yes that is correct