# Why is $|z|^2 = z z^*$?

I've been working with this identity but I never gave it much thought. Why is $|z|^2 = z z^*$ ? Is this a definition or is there a formal proof?

• what is the $z^*$ ? – Mher May 9 '14 at 16:20
• Why do you aks this? Do you know what's the "usual" representation of a complex number as $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ ? Do you know what's the conjugate of a complex number? Then de the multiplication $\;z\overline z\;$ and that's all! – DonAntonio May 9 '14 at 16:21
• @Mher Safaryan It's the complex conjugate. – hb20007 May 9 '14 at 16:24
• @DonAntonio I'm asking because I had no idea it's this simple. The identity was presented to me baldly along with a sloppy explanation to complex numbers back when I took my intro class. Unfortunately, not all educators are able to present mathematics in an accessible, intuitive way. That's a real skill. – hb20007 May 9 '14 at 16:25

I take it that $z^*$ means the conjugate of $z$, then it follows from nothing more than algebra:
$$zz^* = (a+bi) \cdot (a-bi) = a^2 - abi + abi + b^2 = a^2 + b^2 = |z|^2$$
Let $z=x+iy$, for $x,y \in \mathbb{R}$. Then $zz^*=(x+iy)(x-iy)=x^2+y^2=|z|^2$.
Looking at $z=x+yi$ and doing $$zz^*=(x+yi)(x-yi)=x^2+y^2$$ shows that, when we interpret a complex number as a point in the Argand-Gauss plane, $|z|$ represents the distance of the point from the origin.