(My solustion)
Let $k$ be $|ab|$, then $k$ must be one of divisors of $30$.
Suppose $k = 1$, then $ab = e$,
$b = a^{-1}$, but $|a| \ne |b|$.
Hence $k \ne 1$.
Similarly, $2 , 3, 5, 6$ or $10$ cannot be $k$.
If $ k =15$, take $a =$ $e^{{{pi}\over{3}}i}$, $b =$ $e^{{{pi}\over{5}}i}$.
If $ k =30$, take $a =$ $-e^{{{pi}\over{3}}i}$, $b =$ $e^{{{pi}\over{5}}i}$.
Hence the number of order of $|ab|$ can be $15$ and $30$.
I thought the solution looks like incomplete so I asked this question.
I would appreciate if you let me know a better way to solve it. (or there is something wrong)
