Is it valid to multiply both sides of an equation by a complex number? In case I'd like to get rid of some complex denominators, could I multiply both sides of the equation by the common denominator?
I understand there's an other way, but i'd like to know if the above mentioned would be a valid operation, and if not, why.
 A: If $\left(\mbox{some expression}\right)=\left(\mbox{another expression}\right)$, this means that both expressions evaluate to the same complex number. Let's call this number $z$. This means that:
$$z=\left(\mbox{some expression}\right)=\left(\mbox{another expression}\right)$$
Well, if you want to multiply both sides by another complex, say $w$, the equality will still hold, because both $w\cdot\left(\mbox{some expression}\right)$ and $w\cdot\left(\mbox{another expression}\right)$ are just the same as $w\cdot z$.
Beware that if $w$ is $0$ (that is, if you want to multiply both sides by zero), then the equality will still hold: it will be just $0=0$. Of course, this is not useful---you lost all the information you had in your original equation, because you can't divide both sizes by $0$ to get back where you started.
A: It is perfectly valid as long as the multiplier you use isn't zero.
It's even valid to multiply both sides by a variable expression, but again you must include the proviso that variable multiplier cannot be zero. Many are under the mistaken impression that this is not valid, but it is -- you just have to include the additional assumption to the end.
For example, suppose you have the equation $x=1$. It is perfectly valid to multiply both sides by $x$, with the proviso that $x\neq 0$:
$$x=1$$
$$x(x)=x(1)\tag{$x\neq 0$}$$
$$x^2=x\tag{$x\neq 0$}$$
$$x^2-x = 0\tag{$x\neq 0$}$$
$$x(x-1)=0\tag{$x\neq 0$}$$
This equation appears to have an additional solution -- but don't forget that $x=0$ is not a solution since it is excluded by the proviso.
A: This is valid. You can still multiply both sides of an equation by a scalar and still have it hold. It's just like the real numbers.
