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Section 7.3 Geometry and arithmetic of complex numbers

Complex numbers have different representations, and we will exploit this fact regularly:

  1. as a single number \(z\text{,}\)

  2. as a sum of real and imaginary parts \(z=a+ib\text{,}\) and

  3. using geometry, as a point in the complex plane.

Think about the geometric effect of multiplying \(z=x+iy\) by \(i\text{.}\)

\begin{equation*} i(x+iy)=-y+ix \end{equation*}

The geometric transformation that multiplication by \(i\) creates is an anti-clockwise rotation by \(\frac{\pi}{2}\text{.}\)

More generally multiplication of complex numbers can be understood by writing the two numbers in polar form. Let \(z=r(\cos \theta + i \sin \theta)\) and \(w=s(\cos \phi + i \sin \phi)\) then \(zw\)

\begin{align*} zw \amp = r(\cos \theta + i \sin \theta)\times s(\cos \phi + i \sin \phi)\\ \amp = rs ( \cos \theta \cos \phi - \sin \theta \sin \phi)+ rsi (\cos \theta \sin \phi + \sin \theta \cos \phi)\\ \amp = rs (\cos(\theta + \phi) + i\sin(\theta + \phi))\text{.} \end{align*}

Notice the overall geometric effect is to just multiply the moduli and add the arguments.

Since \(i=cos(\pi/2)+i\sin(pi/2)\text{,}\) multiplying \(z\) by \(i\) corresponds to rotating counter clockwise by \(\frac{\pi}{2}\text{,}\) and multiplying by \(-1=cos(\pi)+i\sin(pi)\) corresponds to rotating \(z\) 180 degrees, i.e. flipping \(z\) in the opposite direction.

Since \(z^n\) is \(z\) multiplied together \(n\) times we can prove De Moivre's Theorem.

We prove the first statement by induction. Consider the claim

\begin{equation*} z^n = r^{n} (\cos (n\theta) + i \sin (n\theta)) \end{equation*}

for \(n\) a natural number.

The base case \(n=1\) holds by definition.

Assume our claim holds for some integer \(n\geq 1\text{.}\)

\begin{equation*} z^{n+1} = zz^n = z\times r^{n} (\cos (n\theta) + i \sin (n\theta)) \end{equation*}

using the induction hypothesis. And so

\begin{equation*} z^{n+1}=r(\cos \theta + i \sin \theta) r^{n} (\cos (n\theta) + i \sin (n\theta)) \end{equation*}
\begin{equation*} = r^{n+1} ( \cos \theta\cos (n\theta) - \sin \theta\sin (n\theta) +i(\cos \theta\sin (n\theta) + \sin \theta\cos (n\theta))) \end{equation*}
\begin{equation*} = r^{n+1} (\cos ((n+1)\theta) + i \sin ((n+1)\theta))\text{.} \end{equation*}

The first equation in the theorem holds by the principle of mathematical induction.

For the second part, we just note that,

\begin{equation*} \frac{1}{\cos n \theta +i\sin n \theta} =\cos n \theta - i\sin n \theta\text{,} \end{equation*}

and so

\begin{equation*} z^{-n}=(z^{n})^{-1} = (r^{n}(\cos n\theta + i\sin n\theta))^{-1} = r^{-n} (\cos n \theta - i\sin n \theta)\text{.} \end{equation*}