Suppose $a,b \in \mathbb C$ (with non-zero imaginary parts) satisfy that $a\cdot b \in \mathbb R$. Can I then conclude that $b = \bar a$? Suppose $a,b \in \mathbb C$ (with non-zero imaginary parts) satisfy that $a\cdot b \in \mathbb R$. Can I then conclude that $b = \bar a$?
Can I then conclude that $b = \bar a$, i.e. $b$ is the complex conjugate of $a$ ?
I write $(a_1 + ib_1)(a_2 + ib_2) = (a_1a_2 - b_1 b_2) + i(a_1 b_2 + a_2 b_1)$.
I see that the complex conjugate satisfy that $a_1b_1 + a_2b_1 = 0$, but no the other way around ?
 A: No, you cannot conclude that $b = \bar a$, but you can conclude that $b$ is a real multiple of $\bar a$.
Indeed, if $ab = r \in \mathbb R$, then $\bar a a b = r \bar a  \in \mathbb R \bar a$ and so $b \in \mathbb R \bar a$ since $\bar a a \in \mathbb R$.
The reciprocal is also true, by reading the argument above backwards.
A: No: if you write $a=r e^{i\theta}$ and $b =s e^{i\phi} $ in polar form, it is easy to see what this means:
$$ ab = rs e^{i(\theta+\phi)}. $$
For this to be real, we need $e^{i(\theta+\phi)}=\pm 1$. This occurs if $\theta+\phi$ is a multiple of $\pi$, so you can have $\theta=-\phi$ and $r=s$, which is the complex conjugate, but you can also see that in general this condition just means that the complex numbers have opposite arguments, or rather, that $\bar{a} = kb$ for some real $k$ (which we can see is given by $\bar{a}/b = \frac{r}{s} e^{-i(\theta+\phi)} = \pm r/s$.)
A: No: $a=b=i$ is an easy counterexample.
This would not be true even under the assumption that $|a|=|b|=1$. In general, since $a$ and $b$ are assumed non real, hence non zero, set
$$
u=\frac{a}{|a|}, \quad v=\frac{b}{|b|}
$$
so the other assumption is that $uv=\frac{ab}{|a|\,|b|}$ is real. Since $|uv|=1$, this implies $uv=1$ or $uv=-1$, that is. $v=\bar{u}$ or $v=-\bar{u}$.
