Division of Complex Numbers in Polar Form

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Theorem

Let $z_1 := \polar {r_1, \theta_1}$ and $z_2 := \polar {r_2, \theta_2}$ be complex numbers expressed in polar form, such that $z_2 \ne 0$.


Then:

$\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} \paren {\map \cos {\theta_1 - \theta_2} + i \map \sin {\theta_1 - \theta_2} }$

or:

$\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} \map \cis {\theta_1 - \theta_2}$


Proof

\(\ds \frac {z_1} {z_2}\) \(=\) \(\ds \frac {r_1 \paren {\cos \theta_1 + i \sin \theta_1} } {r_2 \paren {\cos \theta_2 + i \sin \theta_2} }\) Definition of Polar Form of Complex Number
\(\ds \) \(=\) \(\ds \frac {\paren {r_1 \paren {\cos \theta_1 + i \sin \theta_1} } \paren {r_2 \paren {\cos \theta_2 - i \sin \theta_2} } } {\paren {r_2 \paren {\cos \theta_2 + i \sin \theta_2} } \paren {r_2 \paren {\cos \theta_2 - i \sin \theta_2} } }\) multiplying numerator and denominator by $r_2 \paren {\cos \theta_1 - i \sin \theta_1}$
\(\ds \) \(=\) \(\ds \frac {r_1 r_2 \paren {\map \cos {\theta_1 - \theta_2} + i \map \sin {\theta_1 - \theta_2} } } {r_2^2 \paren {\map \cos {\theta_2 - \theta_2} + i \map \sin {\theta_2 - \theta_2} } }\) Product of Complex Numbers in Polar Form
\(\ds \) \(=\) \(\ds \frac {r_1 \paren {\map \cos {\theta_1 - \theta_2} + i \map \sin {\theta_1 - \theta_2} } } {r_2 \paren {\cos 0 + i \sin 0} }\)
\(\ds \) \(=\) \(\ds \frac {r_1 \paren {\map \cos {\theta_1 - \theta_2} + i \map \sin {\theta_1 - \theta_2} } } {r_2 \paren {1 + 0} }\) Cosine of $0 \degrees$ and Sine of $0 \degrees$
\(\ds \) \(=\) \(\ds \frac {r_1} {r_2} \paren {\map \cos {\theta_1 - \theta_2} + i \map \sin {\theta_1 - \theta_2} }\)

$\blacksquare$


Examples

Example: $\dfrac {\paren {2 \cis 15 \degrees}^7} {\paren {4 \cis 45 \degrees}^3}$

$\dfrac {\paren {2 \cis 15 \degrees}^7} {\paren {4 \cis 45 \degrees}^3} = \sqrt 3 - i$


Example: $\dfrac {\paren {8 \cis 40 \degrees}^3} {\paren {2 \cis 60 \degrees}^4}$

$\dfrac {\paren {8 \cis 40 \degrees}^3} {\paren {2 \cis 60 \degrees}^4} = -16 - 16 \sqrt 3 i$


Sources

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