1. ## complex manipulation?????

Q show that 1/(1-e^jx) = 1/2(1+jcot(x/2)) where x=theta,sorry bwt x

the first line of woking is to answer ( e^-jx/2 ) / ( (e^-jx/2 - e^jx/2 )

but how does 1 = e^-jx/2

could someone show me how to get ans pls???

2. Two handy identites to know are:

$sin{\theta}=\frac{e^{i{\theta}}-e^{-i{\theta}}}{2i}$

and $cos{\theta}=\frac{e^{i{\theta}}+e^{-i{\theta}}}{2}$

$e^{i{\theta}}=cos{\theta}+isin{\theta}$

$e^{-i{\theta}}=cos{\theta}-isin{\theta}$

Another one handy to know is the identity $tan(\frac{\theta}{2})=\frac{1-cos{\theta}}{sin{\theta}}$

So, you have $\frac{1}{1-e^{-i{\theta}}}=\frac{1}{1-cos{\theta}+isin{\theta}}$