A hypothesis concerning $\Re(z) $ and $\Re(z^n) $.

Question

I was working on a problem in my textbook until I stumbled upon this little question to which I couldn't find the answer in my textbook (I'm in highschool):

Question: If $\Re(z)=a$ such that $a \in \Bbb R $, can we say that $\Re(z^n)=a^n$ such that $n \in \Bbb N $?

What I think is we should study the sign of $a$ (whether it's negative or positive or equal to 0) and also whether $n$ is odd or even. Meaning :

If $ a>0 $ then $$\Re(z^n) = a^n $$

If $ a<0 $ and $n=2k+1$ such that $ k \in \Bbb Z $ , then $$\Re(z^n) = a^n $$

If $a<0$ and $n=2k$ such that $ k\in \Bbb Z $ , then $$\Re(z^n) = - a^n $$

My understanding is that we should preserve the sign of $a$ when we move from $z$ to $z^n$ , is this correct ?

Answer 1

$Re(i)=0$ but $Re(i^4)=1\not = 0^4$

What is true is $|z^n|=|z|^n$ but that is the magnitude rather than the real part and is always non-negative

Answer 2

Hypothesis: $\Re(z) =a\implies \Re(z^n) =a^n$

Counterexample: $$z =2+i \,, \, \Re(z) =2$$ $$z^2=3+2i\,,\,\Re(z^\color{red} 2)=3 \neq (\Re(z))^{\color{red} 2} =4$$