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I'm trying to find a power series for the following expression: $\frac{1+iz}{1-iz}$. I think I could write this as $$1+iz \,\cdot \frac{1}{1-iz}=(1+iz) \sum_{n=0}^\infty (iz)^n=\sum_{n=0}^\infty (iz)^n+(iz)^{n+1}$$

Is this correct though?

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The calculations are indeed correct (a couple of parantheses wouldn't hurt in the last expression), but If you want to show it in power series form then you should group terms with the same exponent. Do you see why you should get $$ 1+\sum_{n=1}^\infty 2(iz)^n \text{ ?} $$

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  • $\begingroup$ No, I can't see how you found that expression. Could you explain this? $\endgroup$ – MHW Nov 25 '16 at 14:48
  • $\begingroup$ I think the series notation is messing with you; notice that $(1+iz)\sum_{n=0}^\infty(iz)^n=(1+iz+(iz)^2+(iz)^3+\cdots)+(iz+(iz)^2+(iz)^3+\cdots)$. $\endgroup$ – user378947 Nov 25 '16 at 14:53
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    $\begingroup$ Oh I see. I agree, I think it was the notation that was causing the confusion. $\endgroup$ – MHW Nov 25 '16 at 14:57
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A general method:

$${1+iz\over 1-iz}={1\over 1+z^2}(1-z^2+2iz)$$

Let $\cos\theta ={1-z^2\over 1+z^2}\implies \sin\theta ={2z\over 1+z^2}$ So find out $\theta$

Now $$e^{i\theta }=\sum_{j=0}^\infty{(i\theta)^j\over j!}$$

There you get a power series

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