I was wondering, if $\frac{x}{0}$ is undefined, is $\frac{x}{i}$ undefined too?
Considering that the real part of $i$ is zero, is the $\frac{x}{0+i}$ some form of division $x$ by $0$?
Does $\frac{x}{0+i}$ cause any problem of the kind $\frac{x}{0}$ does?
On
If $c\ne 0$ and $mn = c\ne$ then $\frac cm = n$ with no error.
If $m=0$ then $mn =0\cdot n = c\ne 0$ will never happen. So $\frac cm$ is an error.
But if $Re(m) =0$ .... so what? There's no reason that that will cause $mn =c$ to be a problem.
Cosider $a + bi$ then $(a + bi)*i = ai + bi^2 = ai +b(-1) = -b + ai$ . So $\frac {-b+ai}i = a+ bi$ is a perfectly acceptable and reasonable result.
But $Re(i) = 0$.... well, so what? That in no way forces $w*i$ to be equal to $0$. So if $w*i = x$ and $x\ne 0$ there is NOTHING wrong with saying $w = \frac xi$.
.....
In fact....... $(-i)i = -(i*i) =-(i^2) = -(-1) = 1$ So $-i = \frac 1i$.
So If we ever have a $\frac xi$ that is: $\frac xi = -xi$
First note that complex numbers of the form $a+bi$ (where $a$ and $b$ are real numbers) include all the real numbers ($b=0$) and all the purely imaginary numbers ($a=0$). (This also means that $0$ is the only number that is both real and imaginary.)
Any complex number of the form $a+bi$ except for $0$ has a multiplicative inverse (in other words, its reciprocal exists) and therefore dividing by that number is the same as multiplying by the inverse.
Here is an elementary proof for how to explicitly find the multiplicative inverse of $a+bi$:
$$ \frac{1}{a+bi} \frac{a-bi}{a-bi} = \frac{a-bi}{a^2+b^2} $$ Notice that by multiplying the top and bottom of the fraction by the same $a-bi$ expression (this is known as the conjugate of $a+bi$), the denominator now becomes a simple real number, which you can distribute through the $a-bi$ on the top.
This trick doesn't work on $0$ though, as its conjugate $a-bi$ is still $0$, so you would be multiplying your fraction by $ \frac{0}{0}$, which is clearly illegal!