Find the $n$-th power of complex number

Question

Let $z=1+2i$ be a complex number.

Prove that for any $n \in \mathbb{N}^*$, the number $z^n$ has the following form: $a_n+ib_n$, with $a_n,b_n \in \mathbb{Z}$.

I guess the solution lies in the trigonometric form of $z$. However, I get a really ugly number trying to transform $z$ into its trigonometric form.

I would really appreciate if I would get some help on this problem.

Answer 1

You can try it by induction:

For the base case is trivial that $1,2\in \mathbb{Z}$. Then suppose that for $k$ you have that $a_{k},b_{k}\in \mathbb{Z}$. Finally, to prove the last step, check that

$$(1+2i)(a_{k}+ib_k)=(a_k-2b_k)+i(b_k+2a_k)$$

So $a_{k+1}=a_k-2b_k$ and $b_{k+1}=b_k+2a_k$ where both are integers.

Answer 2

All you need to prove is that complex numbers with integer real and imaginary parts are closed under multiplication and the result follows from there. i.e. if $a,b,c,d$ are integers then:

$$(a+bi)(c+di)$$

Forms a complex number with integer real and imaginary parts.