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If I have to verify this identity with the complex numbers... $$\frac{(1+i)^n}{(1-i)^{n-2}}=2i^{n-1}, \quad n\in\Bbb N$$ considering that $n\in\Bbb N$, I can use the principle of induction. I don't think it's complicated playing on the various powers if $n\geq 2$.

But if $n=0, n=1$ I'd have to do the ratio of complex numbers and I certainly can't multiply it by a cross. Any answer is always welcome.

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  • $\begingroup$ @downvoter: Is there a reason for a fast downvoted? $\endgroup$ – Sebastiano Aug 8 at 22:49
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We have the following equalities: $$ \frac{1+i}{1-i}=i\\ (1-i)^2=-2i $$ Raise the first equality to the $n$th power, multiply by the second one, and you're basically there.

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  • $\begingroup$ Are you using the induction principle? Excuse me, but it's nighttime here in Italy and my eyes are closed. :-( Could you please not go fast and put in a few extra steps? $\endgroup$ – Sebastiano Aug 8 at 23:01
  • $\begingroup$ @Sebastiano No induction here. And I really am not going very fast (although I am using few words). Confirm that my two equalities hold. Follow the steps I describe below the equalities. See what you get. Or let me know which steps are giving you trouble. $\endgroup$ – Arthur Aug 8 at 23:19
  • $\begingroup$ Again thank you. I thinked that being $n\in \Bbb N$ I could to use the induction. My trouble are: if I start from the formula if $n=0$ I have: $\frac{(1+i)^0}{(1-i)^{-2}}=2i^{-1}$ and then is there the identity? I have not done the calculus. Same for $n=1$. I have not undertstood your start to prove the identity. $\endgroup$ – Sebastiano Aug 9 at 8:53
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    $\begingroup$ @Sebastiano If $\frac{1+i}{1-i}=i$, then we also have $$\frac{(1+i)^n}{(1-i)^n}=\left(\frac{1+i}{1-i}\right)^n=i^n$$ However, you need the exponent $n-2$ in the denominator. That can easily be fixed by multiplying by $(1-i)^2$: $$\frac{(1+i)^n}{(1-i)^{n-2}}=(1-i)^2\left(\frac{1+i}{1-i}\right)^n=-2i\cdot i^n$$ And that's it. $\endgroup$ – Arthur Aug 9 at 11:01
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Use magnitude and argument.

$\left|\frac{(1+i)^{n}}{(1-i)^{n-2}}\right|=2$

$\arg{\left(\frac{(1+i)^{n}}{(1-i)^{n-2}}\right)}=(n-1)\frac{\pi}{2}$

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  • $\begingroup$ I haven't understood it yet...but I vote for it :-) $\endgroup$ – Sebastiano Aug 8 at 22:58
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    $\begingroup$ @Sebastiano try to read about polar form of complex numbers. They have interesting properties and a lot of times more useful than breaking complex number into real and imaginary part and use normal algebra $\endgroup$ – Rezha Adrian Tanuharja Aug 8 at 23:59
  • $\begingroup$ @Sebastiano oh and, the expression is not only valid for natural numbers but also for real numbers $\endgroup$ – Rezha Adrian Tanuharja Aug 9 at 0:03
  • $\begingroup$ Do I use $e^{i\theta}=\cos \theta+i\sin\theta$? It was an exercise of a book where there are written $n\in\Bbb N$.Thank you very much. $\endgroup$ – Sebastiano Aug 9 at 8:49
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    $\begingroup$ @Sebastiano $Ce^{i\theta}=C(\cos{(\theta)}+i\sin{(\theta)})$ is better. $\left|Ce^{i\theta}\right|=C$ and $\arg{\left(Ce^{i\theta}\right)}=\theta$. When you multiply two complex number, their magnitude is the product of their respective magnitude while their argument is the sum of their respective argument $\endgroup$ – Rezha Adrian Tanuharja Aug 9 at 10:02

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