complex exponentials

**earthboy** · January 26th 2013, 01:33 AM

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

**Prove It** · January 26th 2013, 01:38 AM

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

**earthboy** · January 26th 2013, 01:46 AM

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...

**Prove It** · January 26th 2013, 01:48 AM

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.

**earthboy** · January 26th 2013, 02:27 AM

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..

**Prove It** · January 26th 2013, 02:40 AM

Yes that is correct

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