Polar and Euler forms
Polar and Euler forms
For \(z\ne0\), let \(r=|z|>0\) and let \(\theta\) be an argument in radians. The same number has three useful representations.
| Representation | Form | Conversion |
|---|---|---|
| Cartesian | \(z=x+yi\) | \(x=r\cos\theta,\ y=r\sin\theta\) |
| Polar | \(z=r(\cos\theta+i\sin\theta)\) | \(r=\sqrt{x^2+y^2}\); choose the correct quadrant. |
| Euler | \(z=r\mathrm e^{i\theta}\) | \(\mathrm e^{i\theta}=\cos\theta+i\sin\theta\) |
\(\operatorname{cis}\theta\) abbreviates \(\cos\theta+i\sin\theta\). Arguments differing by \(2k\pi\), \(k\in\mathbb Z\), describe the same point. Zero has modulus zero and no argument.
| Operation | Rule | Geometry |
|---|---|---|
| Product | \(z_1z_2=r_1r_2\mathrm e^{i(\theta_1+\theta_2)}\) | Multiply lengths; add angles. |
| Quotient | \(\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\mathrm e^{i(\theta_1-\theta_2)}\) | Divide lengths; subtract angles. |
| Sum or difference | Convert to Cartesian form first. | Add or subtract the corresponding vectors. |
Example
\(-\sqrt3+i=2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)=2\mathrm e^{\frac{5\pi i}{6}}\).
For \(z_1=2\mathrm e^{\frac{5\pi i}{6}}\) and \(z_2=3\mathrm e^{\frac{\pi i}{3}}\), \(z_1z_2=6\mathrm e^{\frac{7\pi i}{6}}\). Its principal argument is \(-\dfrac{5\pi}{6}\), not \(\dfrac{7\pi}{6}\).
The product rule follows from \(\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta\) and \(\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\): expanding \((\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta)\) gives \(\cos(\alpha+\beta)+i\sin(\alpha+\beta)\).
On the unit circle, \(z=\mathrm e^{i\theta}\) has coordinates \((\cos\theta,\sin\theta)\), satisfying \(x^2+y^2=1\). Increasing \(\theta\) moves anticlockwise.
For nonzero \(z=r\mathrm e^{i\theta}\), \(\dfrac1z=\dfrac1r\mathrm e^{-i\theta}\): reciprocal modulus and reflection in the real axis. When \(|z|=1\), \(z^{-1}=z^*\).
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