Question

For any integer $n$, find the modulus of $z=\frac{(\sqrt{3}+i{)}^{4n+1}}{(1-i\sqrt{3}{)}^{4n}}$.

Answer 1

$z=\frac{{(\sqrt{3}+i)}^{4n+1}}{{(1-i\sqrt{3})}^{4n}}=\frac{{(\sqrt{3}+i)}^{4n+1}{(1-i\sqrt{3})}^{4n}}{{(1-i\sqrt{3})}^{4n}{(1+i\sqrt{3})}^{4n}}=\frac{(\sqrt{3}+i){(\sqrt{3}+i)}^{4n}{(1-i\sqrt{3})}^{4n}}{{(1-{\left(i\sqrt{3}\right)}^2)}^{4n}}=\frac{(\sqrt{3}+i){(\sqrt{3}+3i+i+i^2\sqrt{3})}^{4n}}{{(1+3)}^{4n}}=(\sqrt{3}+i)\frac{{(\sqrt{3}+4i+\sqrt{3})}^{4n}}{{\left(4\right)}^{4n}}=(\sqrt{3}+i)\frac{{\left(4\right)}^{4n}}{{\left(4\right)}^{4n}}{i}^{4n}{i}^4=1$

So $z=\sqrt{3}+i$

$|z|$ = ${\sqrt{3}}^2+1^2=4$