Polar Representation with Multiplication of Complex Numbers

Polar Representation with Multiplication of Complex Numbers

Recall from The Polar Representation of Complex Numbers page that if $z = a + bi \in \mathbb{C}$, then we can consider this number as a vector in the complex plane, and the argument of $z$ is defined to be the angle $\theta$ denoted $\arg (z) = \theta$ formed from this vector with the positive real axis. If we let $r = \mid z \mid$, then we used basic trigonometry to obtain a polar representation $(r, \theta)$ of the complex number $z$ explicitly as:

\begin{align} \quad z = r (\cos \theta + i \sin \theta) \end{align}

We will now see that this polar representation of complex numbers gives us a very nice interpretation of the operation of complex number multiplication.

Theorem 1: Let $z, w \in \mathbb{C}$ and $z, w \neq 0$. Then:
a) $\mid z \cdot w \mid = \mid z \mid \mid w \mid$.
b) $\arg (z \cdot w ) = \arg (z) + \arg (w) \pmod {2\pi}$

The notation "$\pmod 2\pi$" above means that $\arg (z \cdot w) = \arg (z) + \arg (w) + 2k\pi$ for some $k \in \mathbb{Z}$.

  • Proof of a) and b): Let $z, w \in \mathbb{C}$, and let $r_z = \mid z \mid$, $r_w = \mid w \mid$, $\theta_z = \arg (z)$, and $\theta_w = \arg(w)$. Then the polar representations of $z$ and $w$ are:
\begin{align} \quad z = r_z (\cos \theta_z + i \sin \theta_z) \quad \mathrm{and} \quad w = r_w (\cos \theta_w + i \sin \theta_w) \end{align}
  • Taking the product $z \cdot w$ yields:
\begin{align} \quad z \cdot w &= r_z (\cos \theta_z + i \sin \theta_z) \cdot r_w (\cos \theta_w + i \sin \theta_w) \\ \quad &= [r_zr_w][ \cos \theta_z \cos \theta_w - \sin \theta_w \sin \theta_w + i(\cos \theta_z \sin \theta_w + \sin \theta_z \cos \theta_w)] \end{align}
  • Recall the following trigonometric identities:
\begin{align} \quad \cos (u + v) = \cos u \cos v - \sin u \sin v \quad (*) \end{align}
\begin{align} \quad \sin (u + v) = \sin u \cos v + \cos u \sin v \quad (**) \end{align}
  • Using both $(*)$ and $(**)$ give us that:
\begin{align} \quad z \cdot w = [r_z r_w] (\cos (\theta_z + \theta_w) + i \sin(\theta_z + \theta_w)) \end{align}
  • Notice that this complex number is polar form, thus, $\mid z \cdot w \mid = r_zr_w = \mid z \mid \mid w \mid$ and $\arg (z \cdot w) = \theta_z + \theta_w = \arg(z) + \arg(w)$. $\blacksquare$

Geometrically, Theorem 1 tells us that if we multiply two complex numbers $z$ and $w$ then the product $z \cdot w$ is the complex number whose modulus/length is the product of the moduli of $z$ with $w$ and whose argument is the sum of the arguments of $z$ and $w$ as pictured below:

Corollary 1: Let $z, w \in \mathbb{C}$ and let $z, w \neq 0$. Then $\displaystyle{\biggr \lvert \frac{z}{w} \biggr \rvert = \frac{\mid z \mid}{\mid w \mid}}$.
  • Proof: By Theorem 1 we have that:
\begin{align} \quad \biggr \lvert \frac{z}{w} \biggr \rvert \mid w \mid &= \biggr \lvert \frac{z}{w} \cdot w \biggr \rvert \\ &= \mid z \mid \end{align}
  • Therefore $\displaystyle{\biggr \lvert \frac{z}{w} \biggr \rvert = \frac{\mid z \mid}{\mid w \mid}}$. $\blacksquare$
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