If $a \in \mathbb{C}$, is $|a|^2=\bar{a}a=a\bar{a} \in \mathbb{R}$?

Question

If $a \in \mathbb{C}$, is $|a|^2=\bar{a}a=a\bar{a} \in \mathbb{R}$?

Meaning, if I have a complex number and I multiply it by its complex conjugate, would that always return a number in $\mathbb{R}$?

Thanks!

Answer 1

Yes, because if $a = u+i v$, then $a \bar{a} =(u+i v)(u-i v) =u^2 + v^2 = |a|^2 $.

Answer 2

Yes, any complex number $g$ can be written as $g=a+ib$ where $a$ and $b$ are real numbers and $i$ is the imaginary unit. The complex conjugate of $g$ is $\bar{g}=a-ib.$ So $$\bar{g}g=(a+ib)(a-ib)=a^2 + b^2\in \mathbb{R}.$$

Answer 3

One could also look at the polar representation $$z=re^{i\theta}.$$

The conjugate is $\overline z=re^{-i\theta}$.

Hence $z\overline z=re^{i\theta}\cdot re^{-i\theta}=r^2e^{i\theta-i\theta}=r^2\cdot e^0$.

Answer 4

Yes. Let $b=a\bar a$. Then $\bar b = \bar a \bar{\bar a}=\bar a a = b$ and so $b$ is real.