# Verify that $\frac{(1+i)^n}{(1-i)^{n-2}}=2i^{n-1}$ for $n\in\Bbb N$ in the complex numbers

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.

• @downvoter: Is there a reason for a fast downvoted? – Sebastiano Aug 8 at 22:49

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.

• 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? – Sebastiano Aug 8 at 23:01
• @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. – Arthur Aug 8 at 23:19
• 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. – Sebastiano Aug 9 at 8:53
• @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. – Arthur Aug 9 at 11:01

Use magnitude and argument.

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

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

• I haven't understood it yet...but I vote for it :-) – Sebastiano Aug 8 at 22:58
• @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 – Rezha Adrian Tanuharja Aug 8 at 23:59
• @Sebastiano oh and, the expression is not only valid for natural numbers but also for real numbers – Rezha Adrian Tanuharja Aug 9 at 0:03
• 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. – Sebastiano Aug 9 at 8:49
• @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 – Rezha Adrian Tanuharja Aug 9 at 10:02