# Homework Help: Complex number simplification

1. Sep 23, 2010

### danik_ejik

Hello,
I need to simplify and find Re and Im of .

I've multiplied by and got that Re= and Im= .

Is that the correct way and a correct result ?

2. Sep 23, 2010

### sachinism

seems fair to me

3. Sep 24, 2010

### ehild

Show your work in detail. The final result is wrong. How did you get cos x in the denominator?

ehild

4. Sep 24, 2010

### danik_ejik

here's my work in details,
http://img811.imageshack.us/img811/7444/exsol.jpg" [Broken]

Last edited by a moderator: May 4, 2017
5. Sep 24, 2010

### ehild

It is better if you type in your work.

I found a typo: You wrote that e-ix=cos (2x)-isin(2x).

ehild

6. Sep 24, 2010

### danik_ejik

it should have been
[URL]http://latex.codecogs.com/gif.latex?e^{-i2x}[/URL]
so the latter expansion is correct.

Last edited by a moderator: Apr 25, 2017
7. Sep 24, 2010

### ehild

OK, I see now. But you should exclude x=(2k+1)pi/2 before multiplying both numerator and denominator by 1+exp(-2ix) which is 0 if x=(2k+1)pi/2.

8. Sep 24, 2010

### danik_ejik

$$\frac{1+e^{ix}}{1+e^{2ix}}=\frac{1+e^{ix}}{e^{ix}(e^{-ix}+e^{ix})}=\frac{e^{-ix}(1+e^{ix})}{2\cos{x}}$$