How to Express \(\frac{1}{e^{z}-1}\) in \(u+iv\) Form?

[unchained1978]

This is related to another post of mine. How would you go about writing \frac{1}{e^{z}-1} in u+iv form? Usually multiplying through with the complex conjugate gives you the desired form, but here I'm not sure that it works. Any suggestions?

 

[haruspex]

Write z = x+iy. Massage the denominator into u+iv form, then multiply top and bottom by conjugate.

 

[DonAntonio]

This is related to another post of mine. How would you go about writing \frac{1}{e^{z}-1} in u+iv form? Usually multiplying through with the complex conjugate gives you the desired form, but here I'm not sure that it works. Any suggestions?

With \,z=x+iy\,\,,\,x,y,\in \Bbb R\,\,\,and\,\,\,e^{ix}=\cos x+i\sin x :

\frac{1}{e^z-1}=\frac{1}{e^x\cos y -1+ie^x\sin y}=\frac{e^x\cos y-1-e^xi\sin y}{e^{2x}-2e^x\cos y+1}=\frac{e^x\cos y-1}{e^{2x}-2e^x\cos y+1}-\frac{e^x\sin y}{e^{2x}-2e^x\cos y+1}\,i

Piece of cake (pant,pant!)

DonAntonio

 

[Mark44]

This is related to another post of mine. How would you go about writing \frac{1}{e^{z}-1} in u+iv form? Usually multiplying through with the complex conjugate gives you the desired form, but here I'm not sure that it works. Any suggestions?

Write e^z as e^{x + iy} = e^xe^iy = e^x(cosy + i siny). Your denominator is this expression, minus 1.

It's slightly messy, but you can rationalize the denominator by multiplying by the conjugate over itself.