# Help with Eulers relation in Fourier analysis

1. Apr 14, 2013

### Huumah

Hi I'm doing fourier analysis in my signals and system course and i'm looking at the solution to one basic problem but I'm having trouble understanding one step

Can anyone explain to me why

becomes

From Eulers formula: http://i.imgur.com/1LtTiKX.png

for example the Cosine in my problem. I thought the "twos" would cancel each other but instead becomes 4 and simular for the sine.

2. Apr 14, 2013

### CompuChip

If
$$\cos z = \frac{e^{jz} + e^{-jz}}{2}$$
then multiply everything by 2 to get
$$2 \cos z = e^{jz} + e^{-jz}$$

Or you can use the identity
$$e^{jz} = \cos(z) + j \sin(z)$$
and also immediately get
$$e^{jz} + e^{-jz} = \cos(z) + j \sin(z) + \cos(z) - j \sin(z) = 2 \cos(z)$$