Complex Numbers Operations

Conversion formulas

	Cartesian	Exponential
	$z=x+iy$	$z=r{e}^{i\theta }$
conjugate	$z^{\ast }=x-iy$	$z^{\ast }=r{e}^{-i\theta }$
modulus	$\Vert z\Vert =\sqrt{z{z}^{\ast }}$	$\Vert z\Vert =r$
conversion	$x=r\cos \theta$, $y=r\sin \theta$	$r=\sqrt{x^2+y^2}$, $\theta =arctan2(y,x)$

E.g.

$z=1+i\sqrt{3}{z}^{\ast }=1-i\sqrt{3}$

$|z|=\sqrt{1^2+{\sqrt{3}}^2}=\sqrt{4}=2$

$\theta =arctan2(2,1)=\frac{\pi }{3}$

$\therefore z=2e^{i\frac{\pi }{3}}$

a) For $z=3-4i$, find $|z|$ and $\theta$ b) Convert $z=-2+2i$ to exponential form c) Convert $z=2e^{i\frac{\pi }{6}}$ to cartesian form

a)

$z^{\ast }=3+4i$

$|z|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

$\theta =arctan2(4,3)=53.13$

b)

$z=-2+2i$

$|z|=\sqrt{8}=2\sqrt{2}=r$

$\theta =\arctan (2,-2)=3\pi /4$

$\therefore -2+2i=z=2\sqrt{2}\cdot e^{i\frac{3\pi }{4}}$

c)

$z=2e^{i\frac{\pi }{6}}r=2,\theta =\frac{\pi }{6}$

$x=2\cos \left(\frac{\pi }{6}\right)=\frac{2\sqrt{3}}{2}=\sqrt{3}$

$y=2\sin \left(\frac{\pi }{6}\right)=\frac{2\cdot 1}{2}=1$

$\therefore z=\sqrt{3}+i$

Basic Operations

$z_1+z_2=(x_1+x_2)+i(y_1+y_2)$

$z_1-z_2=(x_1-x_2)+i(y_1-y_2)$

$z_1\cdot z_2=r_1{e}^{i{\theta }_1}\cdot r_2{e}^{i{\theta }_2}=r_1\cdot r_2{e}^{i({\theta }_1+{\theta }_2)}$

$\frac{z_1}{z_2}=\frac{r_1{e}^{i{\theta }_1}}{r_2{e}^{i{\theta }_2}}=\frac{r_1}{r_2}{e}^{i({\theta }_1-{\theta }_2)}$

$z_1\cdot z_2=(x_1+i{y}_1)(x_2+i{y}_2)=x_1{x}_2-y_1{y}_2+i(x_1{y}_2+x_2{y}_1)$

$\frac{z_1}{z_2}=\frac{(x_1+i{y}_1)(x_2-i{y}_2)}{x_2^2+y_2^2}=\frac{x_1{x}_2+y_1{y}_2}{x_2^2+y_2^2}+i\frac{x_2{y}_1-x_1{y}_2}{x_2^2+y_2^2}$

E.g.

$z_1=\frac{1}{2}+\frac{\sqrt{3}}{2}i=e^{i\frac{\pi }{3}}$

$z_2=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i=e^{i\frac{\pi }{4}}$

$z_1{z}_2=e^{i\left(\frac{\pi }{3}+\frac{\pi }{4}\right)}=e^{i\frac{7\pi }{12}}=\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}$

$\frac{z_1}{z_2}=e^{i\left(\frac{\pi }{3}-\frac{\pi }{4}\right)}=e^{i\frac{\pi }{12}}=\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}$

Or

$z_1{z}_2=\left(\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4}\right)+i\left(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\right)$

$\frac{z_1}{z_2}=\left(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\right)+\left(-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\right)i$

$\cos \frac{7\pi }{12}=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\sin \frac{\pi }{12}=-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}$

Conjugation

$|z^{\ast }|=|z|$

$(z_1\pm z_2{)}^{\ast }=z_1^{\ast }\pm z_2^{\ast }$

$(z_1\cdot z_2{)}^{\ast }=z_1^{\ast }\cdot z_2^{\ast }$

$(z_1/z_2{)}^{\ast }=z_1^{\ast }/z_2^{\ast }$

$(z^x{)}^{\ast }=(z^{\ast }{)}^x(x\in \mathbb{R})$

$(x^z{)}^{\ast }=x^{z^{\ast }}(x\in \mathbb{R})$

Powers and Roots

$z=r{e}^{i\theta }$ and $s$, $s\in \mathbb{R}$, power given by

$z^s=r^s{e}^{is\theta }$

⇒ From this we derive De Moivre’s theorem:

$(\cos \theta +i\sin \theta {)}^s=\cos (s\theta )+i\sin (s\theta )$

Advanced Operations

Evaluate $\sqrt{i}$ (which is $\sqrt{\sqrt{-1}}$)

$\sqrt{i}={\left(e^{\frac{\pi i}{2}}\right)}^{\frac{1}{2}}=e^{\frac{\pi i}{4}}=\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}(1+i)$

$=\frac{1}{2}(1+i{)}^2=\frac{1}{2}(1+2i+i^2)=\frac{1}{2}(2i)=i$

Evaluate ${\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)}^{50}$

${\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)}^{50}\Rightarrow |z|=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=\sqrt{\frac{4}{4}}=\sqrt{1}=1$

$r=1\Rightarrow x=r\cos \theta ,y=r\sin \theta \theta =\arctan \left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)=\frac{\pi }{3}$

$\therefore {\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^{50}={\left(e^{i\frac{\pi }{3}}\right)}^{50}=e^{i\frac{50\pi }{3}}=e^{\left(16+\frac{2}{3}\right)\pi i}$

$=e^{i\frac{2\pi }{3}}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$