Is the Complex Conjugate the Only Way to Get a Real Number?

Question

Is the complex conjugate of a number (or a real multiple of it) the only complex number which, when multiplied with the original number, gives a real number?

Answer 1

Let $z,w \in \mathbb C$ and suppose $zw = k \in \mathbb R\setminus \{0\}$. Without loss of generality, $|w|=1$ so that $\frac1w = \overline w$. Then $$z = \frac kw = k\overline w$$So $w = \dfrac 1k \overline z$

If $zw = 0$ and $z\ne 0$, then $w = 0 = 0\cdot\overline z$.

However, if $z=0$, then $\overline z = 0$. But $zw = 0 \in \mathbb R$ for all $w \in \mathbb C$, and if $w \ne 0$, then $w$ is not a multiple of $\overline z = 0$.

Answer 2

Suppose we are given a complex number $a+bi$ and we wish to find another complex number $c+di$ such that $(a+bi)(c+di)$ is real. Since $(a+bi)(c+di) = (ac-bd)+(ad+bc)i$, we need $ad+bc = 0$. If we know that $b \neq 0$ (i.e. the given number isn't real), then we have $c = -\dfrac{a}{b}d$. Thus, $c+di = -\dfrac{a}{b}d+di = -\dfrac{d}{b}(a-bi)$, which is a real multiple of the conjugate of $a+bi$.

Answer 3

Since Complex multiplication of $z$ by $w:=c+id$ involves rotation by $arg(w)$ , w must have $$argw +argz= k\pi ; k \in \mathbb Z$$

Answer 4

If $z=re^{i\phi}$ and $w=se^{i\psi}$ where $r,s\in [0,\infty)$ and $\phi,\psi\in[0,2\pi)$ then: $$zw=rse^{i(\phi+\psi)}\in \mathbb R\iff rs=0\vee \phi+\psi\in \{0,2\pi\}\iff rs=0\vee w\text{ is multiple of }\overline{z}$$

Answer 5

Various answers have already provided symbolic proofs of the theorem you needed, but if you want a visually intuitive reason, consider:

When you multiply two complex numbers you multiply their distances from the origin and add their angles relative to the real axis. So how can you end up on the real axis? One way is to multiply by the complex conjugate so your angles sum to zero. Every other way is equivalent to multiplying by the complex conjugate first (to get to the real axis), then multiplying by some real number to move back and forth along the real axis.

Answer 6

Let $z$ be you complex number; you want to find all $z'\in\Bbb C$ such that $zz'\in\Bbb R$. There are two relevant cases:

1. $z=0$, for which every $z'\in\Bbb C$ will do.
2. $z\neq 0$ in which case $zz'=r\in\Bbb R$ is equivalent to $z'=\frac rz$ (with still $\in\Bbb R$). Then your solution set is formed by all real multiples of $\frac1z$. Since $\overline z=\frac{|z|^2}z$ is a nonzero real multiple of $\frac1z$, this is the same as the set of all real multiples of $\overline z$.