Two complex numbers have a product of $30$ and positive integers as real parts. Write all possible combinations of such numbers. 
Two complex numbers have a product of $30$ and positive integers as real parts. Write all possible combinations of such numbers.

$(3,10), (5,6),(2,15)$ and $(1+ i\sqrt(29),1-i\sqrt(29)$ are few solutions. Is there any neat way to find all? I tried by assuming numbers to be $m+ia$ and $n+ib$ but this was does not seems neat.
 A: Assume that $n,m$ are positive integers. We have that
$$(m+ia)(n+ib)=mn-ab+i(an+bm)=30$$
implies $an=-bm$, that is $a=-bm/n$. It follows that
$$mn-ab=mn+\underbrace{b^2\frac{m}{n}}_{\geq 0}=30$$
Therefore $mn\le 30$ and $b=\pm\sqrt{\frac{n}{m}(30-mn)}$.
So by choosing all couples of positive integers $(m,n)$ such that $mm\leq 30$ you will find all possible solutions for your problem:
$$m+ia=m\mp i\sqrt{\frac{m}{n}(30-mn)}\quad,\quad
n+ib=n\pm i\sqrt{\frac{n}{m}(30-mn)}.$$
A: If you represent your complex numbers as
$$w=m+iq \,,~\,~\,~\,~ z=n+ir  \,,$$
for natural numbers $m,n$ (and real numbers $q,r$), then your hypotheses equivocates to
$$mn-qr =30\,,\,~\,~\,~ mr + nq =0\,.$$
It follows from the first that $qr$ is an integer, and from the second equation that it is negative or $q=r=0$; multiplying the second through by $q$ and $r$ also shows that $q^2$ and $r^2$ are both rational numbers. Explicitly (solving them algebraically), that means the possible combinations take the form
$$w=m\pm i\sqrt{m^2-30\frac{m}{n}}\,,~\,~\,~\,~ z=n\mp i\sqrt{n^2-30\frac{n}{m}}\,,$$ with the caveat that $mn\le 30$ (and, a fortiori, the imaginary parts of $w$ and $z$ are of opposite signs).
A: Taking norms, we get
$$\sqrt{m^2+a^2}\sqrt{n^2+b^2} = 30$$
where all quantities involved are real. The left hand side is clearly larger than or equal to $\sqrt{m^2}\sqrt{n^2}=mn$ (since the question requires that $m,n>0$). Therefore $30\geq mn$, now there are only a few cases to check.
