Calculate $\frac {(1 +i)^n}{(1-i)^{n-2}}$ where $n$ is an integer such that $n\ge2$

Question

Calculate $\frac {(1 +i)^n}{(1-i)^{n-2}}$ where $n$ is an integer such that $n\ge2$

Evaluating $\frac{(1+i)^{n+2}}{(1-i)^n}$ Is very similar to this one; actually, with the information given in this problem i got that:

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

But evaluating at $n=4$ and $n=5$ the results are different. I’d really appreciate some help. Thanks

Answer 1

Your result is absolutely right

$\frac {(1+ i )^4}{(1-i )^2}=\frac {(2i )^2}{-2i}= \frac{-4}{-2i}=\frac {2}{i} = \frac{2i}{i^2} = -2i$

$-2i^{4+1} =-2i^{5} $

$$i^4 = 1$$

So $-2i^{5}=-2i$

Answer 2

As Octavio Berlanga noted, $\frac {1+ i}{1-i} =\frac {i(1+ i)}{i(1-i)} =\frac {i(1+ i)}{i+1} = i $

so

$\begin{array}\\ \frac {(1 +i)^n}{(1-i)^{n-2}} &=\frac {(1 +i)^{n-2}(1+i)^2}{(1-i)^{n-2}}\\ &=i^{n-2}(1+2i-1)\\ &=i^{n-2}(2i)\\ &=2i^{n-1}\\ &=2(-i, 1, i, -1, \text{repeated starting at 0})\\ \end{array} $