Celeb Glow
general | April 15, 2026

What is the bar indicating in $\overline{z}$?

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I was asked to solve the following question

prove $z=\overline{z}$

conditions : iff $z$ is real.

To begin with I have no idea what $\overline{z}$ is supposed to be what is the bar indicating here. I assumed opposite?

Can anyone give an explanation of how I might go about solving the question. And what the bar is supposed to be ?

(apologies if the question seems novice. I am not exactly a math whiz.)

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2 Answers

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The notation $\bar{z}$ indicates the complex conjugate of $z$, which is defined as $$\bar{z}=a-bi,$$ where $z=a+bi$. In physics, this may be denoted $z^{*}$ instead. Note that if $z$ is real, then $b=0$ and thus $$z=a+0i=a-0i=\bar{z}.$$ Conversely, if $z=\bar{z}$ then $$a+bi=a-bi\Longrightarrow bi=-bi,$$ namely $b=-b$. But, this implies that $b=0$ and moreover, that $z=a+0i=a$ is real. Hence, $$z=\bar{z}\iff z\in\Bbb R.$$


Geometrically, one can picture the complex conjugate as an automorphic map (meaning from $\Bbb C\to\Bbb C$) which reflects $z$ across the real axis. With this interpretation, it makes sense that real numbers (represented as coordinates on the complex plane with a $y$ value of zero) are precisely those that equal their conjugates.

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If $z=a+bi$, by definition $$\overline z=a-bi$$

If $z=\overline z$, $$a+bi=a-bi\implies \color{RED}{b=0}$$

Graphically, considering $z$ as a vector on the complex plane, $\overline z$ is the reflection of $z$ along the real axis!

Brain exercise: what vectors would remain unchanged upon reflection along real axis?

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