Splitting $(\frac{1+i}{1-i})^k$ into real and immaginary parts

Question

I can rewrite the Expression to $(1,1)^k((1,-1)^{-1})^k=(1,1)(\frac{1}{2},\frac{1}{2})^k$.

What can I do now is there a trick I can use with distributivity?

Or do I have to prove something like

$(a,b)^k\cdot (c,d)^k=((a,b)\cdot(c,d))^k\tag{1}$

I am not sure if $(1)$ holds I have started to look at $(a,b)^k$ but it didn't look very promissing

$(a,b)^k=(a^2-b^2,2ab)(a,b)…=(a^3-ab^2-2ab^3,a^2b-b^3+2a^2b)(a,b)...$

Answer 1

Hint:

Get rid of $i$ in the denominator first. That is, you start like so:

$$\frac{1+i}{1-i} = \frac{1+i}{1+i}\cdot \frac{1+i}{1-i} = \cdots$$

Answer 2

The easiest way to solve this kind of problem is to multiply numerator and denominator by denominator's conjugate.

$$\left(\frac{1+i}{1-i}\right)^k=\left(\frac{1+i}{1-i}\times \frac{1+i}{1+i}\right)^k$$ $$=\left(\frac{(1+i)^2}{2}\right)^k=\left(\frac{2i}{2}\right)^k=i^k$$ I hope you can proceed now.