Why is $ |z|^2 = z z^* $?
By Andrew Adams •
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?
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$\begingroup$I take it that $z^*$ means the conjugate of $z$, then it follows from nothing more than algebra:
$$zz^* = (a+bi) \cdot (a-bi) = a^2 - abi + abi + b^2 = a^2 + b^2 = |z|^2$$
$\endgroup$ $\begingroup$Let $z=x+iy$, for $x,y \in \mathbb{R}$. Then $zz^*=(x+iy)(x-iy)=x^2+y^2=|z|^2$.
$\endgroup$ $\begingroup$There is no formal proof: it's a definition.
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.
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