I'm talking about this:
$$c(a+bi)(a-bi) = c(a^2+b^2)$$ where $c$ is real.
Is it true that $a+bi$ can only be multiplied with $c(a-bi)$ in order to get a real number or are there other complex numbers that can multiply to get a real number? If this is the only way, how do you prove it?
On
If $z$ is any nonzero complex number and $w$ is a complex number such that $zw$ is real, then $w$ must be a real multiple of the conjugate $\overline{z}$. The quick way to see this is to notice that both $zw$ and $z\overline{z}$ are real, so the quotient $$\frac{zw}{z\overline{z}}=\frac{w}{\overline{z}}$$ is also real (here we use the assumption that $z$ is nonzero to be sure this division is defined). Since $$w=\frac{w}{\overline{z}}\cdot \overline{z},$$ this says exactly that $w$ is a real multiple of $\overline{z}$.
On
A geometrical way to think about is as follows: Let $z=a+ib=re^{i\alpha}$, you want to multiply this by a complex number $w=se^{i\beta}$ such that the product is real. This would mean $$wz=rse^{i(\alpha+\beta)} \in \Bbb{R}.$$ For this either $s=0$ OR $\alpha+\beta =n\pi \implies \beta=n\pi-\alpha$, where $n \in \Bbb{Z}$. Thus $$w=se^{i(n\pi-\alpha)}=s \, \underbrace{e^{-i\alpha}}_{\frac{\bar{z}}{r}} \, \underbrace{e^{i n\pi}}_{\pm 1}=\pm \frac{s}{r}\bar{z}$$
Well, obviously $0z$ will be real for all $z$. For $w\ne 0$, recall that $\frac1{w}=\frac{\overline w}{\lvert w\rvert^2}$, and therefore $zw=\alpha\in\Bbb R$ implies $z=\frac{\alpha}{\lvert w\rvert^2}\overline w$, which is a real multiple of $\overline w$.