The polar form greatly simplifies the process of complex multiplication and division as compared to
the cartesian form. However, addition and subtraction is a lot more complicated.
For special cases, the following “half angle trick” can be useful, and relies on the conjugate property of
complex numbers.
z + z ∗ = 2 Re ( z ) = 2 r cos θ z + z^* = 2\textrm{Re}(z) = 2 r \cos \theta z + z ∗ = 2 Re ( z ) = 2 r cos θ z − z ∗ = 2 Im ( z ) i = 2 i r sin θ z - z^* = 2\textrm{Im}(z)\mathrm{i} = 2 \mathrm{i} r \sin \theta z − z ∗ = 2 Im ( z ) i = 2 i r sin θ
Examples
Simplify e i θ + 1. {\mathrm{e}^{\mathrm{i}\theta} + 1.} e i θ + 1.
Step 1:
Let e i θ = e i θ 2 e i θ 2 {\mathrm{e}^{\mathrm{i}\theta} = \mathrm{e}^{\mathrm{i}\frac{\theta}{2}}\mathrm{e}^{\mathrm{i}\frac{\theta}{2}}} e i θ = e i 2 θ e i 2 θ 1 = e i θ 2 e − i θ 2 {1 = \mathrm{e}^{\mathrm{i}\frac{\theta}{2}}\mathrm{e}^{-\mathrm{i}\frac{\theta}{2}}} 1 = e i 2 θ e − i 2 θ
e i θ + 1 = e i θ 2 e i θ 2 + e i θ 2 e − i θ 2 = e i θ 2 ( e i θ 2 + e − i θ 2 ) \begin{aligned}
& \mathrm{e}^{\mathrm{i}\theta} + 1 \\
&= \mathrm{e}^{\mathrm{i}\frac{\theta}{2}}\mathrm{e}^{\mathrm{i}\frac{\theta}{2}} +
\mathrm{e}^{\mathrm{i}\frac{\theta}{2}}\mathrm{e}^{-\mathrm{i}\frac{\theta}{2}} \\
&= \mathrm{e}^{\mathrm{i}\frac{\theta}{2}}
\left(\mathrm{e}^{\mathrm{i}\frac{\theta}{2}} + \mathrm{e}^{-\mathrm{i}\frac{\theta}{2}}\right) \\
\end{aligned} e i θ + 1 = e i 2 θ e i 2 θ + e i 2 θ e − i 2 θ = e i 2 θ ( e i 2 θ + e − i 2 θ ) Simplify e i 2 θ − 1. {\mathrm{e}^{\mathrm{i}2\theta} - 1.} e i 2 θ − 1.
Step 1:
Let e i 2 θ = e i θ e i θ {\mathrm{e}^{\mathrm{i}2\theta} = \mathrm{e}^{\mathrm{i}\theta}\mathrm{e}^{\mathrm{i}\theta}} e i 2 θ = e i θ e i θ 1 = e i θ e − i θ {1 = \mathrm{e}^{\mathrm{i}\theta}\mathrm{e}^{-\mathrm{i}\theta}} 1 = e i θ e − i θ
e i 2 θ − 1 = e i θ e i θ − e i θ e − i θ = e i θ ( e i θ − e − i θ ) \begin{aligned}
& \mathrm{e}^{\mathrm{i}2\theta} - 1 \\
&= \mathrm{e}^{\mathrm{i}\theta}\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{\mathrm{i}\theta}\mathrm{e}^{-\mathrm{i}\theta} \\
&= \mathrm{e}^{\mathrm{i}\theta}
\left(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}\right)
\end{aligned} e i 2 θ − 1 = e i θ e i θ − e i θ e − i θ = e i θ ( e i θ − e − i θ )