Polar form arithmetic
Complex numbers in polar form, especially when expressed in exponential form
r e i θ , {r\mathrm{e}^{\mathrm{i}\theta},} r e i θ , lends itself especially well to multiplication, division
and conjugation .
( r 1 e i θ 1 ) ⋅ ( r 2 e i θ 2 ) = ( r 1 r 2 ) e i ( θ 1 + θ 2 ) ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ arg ( z 1 z 2 ) = arg ( z 1 ) + arg ( z 2 ) \begin{aligned}
(r_1\mathrm{e}^{\mathrm{i}\theta_1}) \cdot (r_2\mathrm{e}^{\mathrm{i}\theta_2}) &= (r_1r_2)\mathrm{e}^{\mathrm{i}(\theta_1 + \theta_2)} \\
\left| z_1 z_2 \right| &= \left| z_1 \right| \left| z_2 \right| \\
\arg(z_1 z_2) &= \arg(z_1) + \arg(z_2)
\end{aligned} ( r 1 e i θ 1 ) ⋅ ( r 2 e i θ 2 ) ∣ z 1 z 2 ∣ arg ( z 1 z 2 ) = ( r 1 r 2 ) e i ( θ 1 + θ 2 ) = ∣ z 1 ∣ ∣ z 2 ∣ = arg ( z 1 ) + arg ( z 2 )
r 1 e i θ 1 r 2 e i θ 2 = r 1 r 2 e i ( θ 1 − θ 2 ) ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ arg ( z 1 z 2 ) = arg ( z 1 ) − arg ( z 2 ) {\begin{aligned}
\frac{r_1\mathrm{e}^{\mathrm{i}\theta_1}}{r_2\mathrm{e}^{\mathrm{i}\theta_2}} &= \frac{r_1}{r_2}\mathrm{e}^{\mathrm{i}(\theta_1 - \theta_2)} \\
\left| \frac{z_1}{z_2} \right| &= \frac{\left| z_1 \right|}{\left| z_2 \right|} \\
\arg\left(\frac{z_1}{z_2}\right) &= \arg\left(z_1\right) - \arg\left(z_2\right)
\end{aligned}} r 2 e i θ 2 r 1 e i θ 1 ∣ ∣ z 2 z 1 ∣ ∣ arg ( z 2 z 1 ) = r 2 r 1 e i ( θ 1 − θ 2 ) = ∣ z 2 ∣ ∣ z 1 ∣ = arg ( z 1 ) − arg ( z 2 )
( r e i θ ) ∗ = r e − i θ ∣ z ∗ ∣ = ∣ z ∣ arg ( z ∗ ) = − arg ( z ) \begin{aligned}
(r\mathrm{e}^{\mathrm{i}\theta})^* &= r\mathrm{e}^{-\mathrm{i}\theta} \\
\left| z^* \right| &= \left| z \right| \\
\arg(z^*) &= -\arg(z)
\end{aligned} ( r e i θ ) ∗ ∣ z ∗ ∣ arg ( z ∗ ) = r e − i θ = ∣ z ∣ = − arg ( z )
Principal argument
The problem with angles are that there are more than one way to express the same angle (by going
multiple rounds). For example, the angles 1 4 π \frac{1}{4}\pi 4 1 π represents the same angle as 9 4 π \frac{9}{4}\pi 4 9 π
and − 3 4 π -\frac{3}{4}\pi − 4 3 π .
We resolve this issue by defining the principal argument of a complex number to be between − π -\pi − π and π \pi π .
Principal argument: − π < θ ≤ π . {-\pi < \theta \leq \pi.} − π < θ ≤ π .
If any of our arguments are outside of this range, we can add or subtract 2 k π 2k\pi 2 k π from it to bring it back.
Examples
Evaluate 2 e 2 3 π i × 3 e 5 6 π i {2 \mathrm{e}^{ \frac{2}{3} \pi \mathrm{i} } \times 3 \mathrm{e}^{ \frac{5}{6} \pi \mathrm{i} }} 2 e 3 2 π i × 3 e 6 5 π i
( 2 e 2 3 π i ) ⋅ ( 3 e 5 6 π i ) z = ( 2 ⋅ 3 ) e i ( 2 3 π + 5 6 π ) z = 6 e 3 2 π i
\begin{aligned}
& \left(2 \mathrm{e}^{ \frac{2}{3} \pi \mathrm{i} }\right) \cdot \left(3 \mathrm{e}^{ \frac{5}{6} \pi \mathrm{i} }\right) \\
\phantom{z} &= \left(2\cdot3\right) \mathrm{e}^{\mathrm{i}\left( \frac{2}{3} \pi + \frac{5}{6} \pi \right)} \\
\phantom{z} &= 6\mathrm{e}^{\frac{3}{2} \pi\mathrm{i}}
\end{aligned}
z z ( 2 e 3 2 π i ) ⋅ ( 3 e 6 5 π i ) = ( 2 ⋅ 3 ) e i ( 3 2 π + 6 5 π ) = 6 e 2 3 π i