The Geometry of Complex Number and Its Multiplicative Inverse

Question

I am confused that how a complex number and its multiplicative inverse are related geometrically. For example, the complex conjugate is found by reflecting z across the real axis. Please explain if there is such relationship between a complex number and its multiplicative inverse. It's a problem given in a book authored by Dennis G. Zill.

Answer 1

A complex number $a + bi$ when thought of as a point $(a,b)$ existing in a 2-D plan will have a magnitude $r = \sqrt{a^2 + b^2}$ which is it's distance from the origin, and an incidental angle $\theta$ measuring the angle of the point $(a,b)$ to the origin to the $x$ axis.

As $a$ is the real/$x$ value and $b$ is the imaginary/$y$ value and $r = \sqrt{a^2 + b^2}$ is the radius of a circle around the origin containing $a$ and $b$, we know that $a = r*\cos \theta$ and $b = r*\sin \theta$.

And so if $c + di$ has magnitude $s$ and angle $\eta$ then

$(c + di)(a + bi) = $

$(s*\cos \eta + i s*\sin \eta)(r*\cos \theta + i r*\sin \theta)= $

$rs[(\cos \theta\cos\eta - \sin\theta\sin\eta)+ i(\sin\eta\cos \theta + \sin \theta \cos \eta)]$

=rs(cos (\theta + eta) + i \sin(\theta + eta))$.

This is a rather fascinating result that $(a+bi)(c + di)$ is the result of multiplying their sizes together and adding their angles together.

Is if $z * \frac 1z = 1= 1 + 0i$ (which has size $1$ and incidental angle $0^\circ$) and $z$ has size $r$ and incidental angle $\theta$, then $\frac 1z$ will have size $s= \frac 1r$ because we need $r*s = 1$, and $\frac 1z$ will have an incidental angle of $\eta = -\theta$ because we need $\theta + \eta=0$.

So that's that.

Geometrically, $\frac 1z$ is the complex number that is the inverse of size, and the negative of angle to $z$.

Answer 2

You can have a simple geometric representation using the geometric representation of the complex conjugate and the fact that $$ \frac{1}{z}=\frac{\bar z}{|z|^2} $$

Answer 3

It's actually a fairly simple geometric construction.

Define point $A$ in the complex plane as your complex number, $O$ as zero and $E$ as $1+0i$. ($E$ from German "Einheit" for "unity"). Construct the following two rays:

$OB$ such that $\angle EOB$ is congruent with $\angle EOA$ but on the opposite side of line $OE$.

$EC$ such that $\angle OEC$ is congruent with $\angle OAE$ and on the same side of line $OE$ as ray $OB$ rendered above.

Rays $OB$ and $EC$ intersect at point $Z$ such that $\triangle OZE$ is similar to$\triangle OEA$. This similarity plus the polar form for a complex inverse guarantees that $Z=A^{-1}$.