Find all n for which z is a purely imaginary number

Question

Find every $n\in \Bbb{N}$ for which $z=\dfrac{(1-i)^{3n+4}}{{2^n}(1+i)}$ is a purely imaginary number.

I know $1-i = \sqrt 2\cdot e^{\frac{7\pi}{4}i}$

Then argument of the numerator is

$(3n+4)$ $\cdot \frac{7\pi}{4} +2k\pi \quad k \in \Bbb{Z}$

And I know $1+i= \sqrt 2\cdot e^{\frac{\pi}{4}i}$

So the argument of the denominator is still $\\\frac{\pi}{4}$ (right?) because $2^n$ is just a real number?

And I want the argument of $z$ to be either $\frac{\pi}{2}$ or $\frac{3\pi}{2}$?

Please help me find my mistakes and finish this exercise! Many thanks in advance.

Answer 1

Since $2^n\in\mathbb R$ and since $\dfrac{1-i}{1+i}=-i$, the number $z$ is purely imaginary if and only if $(1-i)^{3n+3}$ is real. Can you take it from here?

Answer 2

HINT: Note that $(1-i)^2=-2i$ and $(1-i)^3=-2(1+i)$.

Answer 3

$\frac{(1-i)^{3n+4}}{{2^n}(1+i)}=\frac{(1-i)^{3n+5}}{2^{n+1}}.$

Hence $\frac{(1-i)^{3n+4}}{{2^n}(1+i)} \in i \mathbb R \iff (1-i)^{3n+5} \in i \mathbb R.$

Now observe that $(1-i)^2= -2i$. Can you procced ?