I need to solve the following problem: Let $z = 5e^{2πi/3}$ . Find the smallest positive integer n such that $(zi)^n ∈ \mathbb{Q}$.
So i equated $(zi)^n$ to $a/b$, where $a$ and $b$ are integers.
Then I have $(zi)^n=(5ie^{2πi/3})^n = a/b$ which $= (5i)^n e^{2nπi/3}$ and $i = e^{π/2 i}$ so $i^n = e^{π/2in}$ so subbing in $i^n$ gives $$5e^{π/2in} e^{2πin/3} = 5e^{7πin/6}$$
This is where i got up to so far, not sure if its correct though. How do i go on from this point?
Really appreciate your help! Thanks!
$i=\cos{\pi/2}+i\sin{\pi/2}=e^{i\pi/2}$
$\displaystyle z=(5ie^{2πi/3})^n=(5e^{i\pi/2}e^{2πi/3})^n= (5e^{(2\pi/3+\pi/2)i})^n =5^ne^{n(2\pi/3+\pi/2)i}=5^n(\cos{(7n\pi/6)}+i\sin{(7n\pi/6)})=a+ib$
If $b=0$ then $(zi)^n ∈ \mathbb{Q}$
$\to 7n\pi/6=k\pi \to n=\dfrac{6k}{7}(n,k ∈ \mathbb{N})$
For $k=7$, $n=6$. This is the first and least integer $n$ that satisfies the conditions.