If $(1-i)^n = 2^n$ , then find $n$.

Question

If $$(1-i)^n = 2^n$$ then find $n$.

If anything raised to $0$ is $1$, but according to my book $ n \ne 0$. Is the print wrong?

Answer 1

It is true only for $n=0$, since $$ \left| (1-i)^n\right|=\big(\sqrt{2}\big)^n=2^{n/2}. $$ Thus, if $(1-i)^n=2^n$, then $2^{n/2}=\left| (1-i)^n\right|=2^n$, and hence $n=0$.

Note. However, what happens more often is $$ (1-i)^{2n}=2^n. $$ Indeed, $(1-i)^{2}=-2i$, and $(1-i)^{4}=-4$, and $(1-i)^8=16$.

In fact $$ (1-i)^{2n}=2^n\quad\text{if and only if}\quad n=4k. $$

Answer 2

$(1-i)^n=2^n$

$(1-i)^n/(1+i)^n\times (1+i)^n =2^n$

$2^n/(1+i)^n=2^n$

$(1+i)^n=1$

$n=0$