$z=x+iy, \quad x, y \in R$

Argand plane or complex plane

$r$ - distance from the origin O to the point P.

$θ$ -angle between OP and the +(ve) x-axis.

$\cos \theta = \frac {x}{r}$

$x = r \cos \theta$

$y = r \sin \theta$

$z = r \cos \theta + i\ r \sin \theta \quad r ≥ 0 , \theta \in [0, { \theta} {\pi}]$

$z = r (cos \theta + i\sin \theta)$

This representation is called the polar representation of $z$.

Polar coordinates system, $(r, \theta)$

Cartesian coordinates system, $(x, y)$

Given $(r, \theta) \mapsto x=r \cos \theta, y=r \sin \theta$

$r = \sqrt {x^2+y^2} \mapsto (x, y)$

$\tan \theta = \frac {y} {x}$

$z = r(\cos \theta + i \sin \theta)$

$z = r$ cis $\theta,\quad$ cis $\theta = \cos \theta +i \sin \theta$

• $\theta$ -argument of $z$
• $r=|z| = \sqrt {x^2+y^2}$

(1) $z=r (\cos\theta + i \sin\theta)$

$z=r [(\cos(\theta +2k\pi) +i \sin(\theta +2k\pi)] \ k\in \Z.$

i.e., $(r, \theta +2k\pi) \mapsto x= r\cos\theta, \\ \quad \quad \quad \quad \quad \quad y=r\sin\theta$

(2) $\tan\theta = \frac{1} {1}$
$\theta = \frac{\pi} {4}$

$z =1+i = {\sqrt 2}\biggl(\cos\frac {\pi} {4}+i \sin\frac {\pi} {4} \biggl)$

consider ${\sqrt 2}\biggl[\cos(\frac {-\pi} {4})+i \sin(\frac {-\pi} {4}) \biggl]$

$= { \sqrt 2}\biggl[\cos\frac {\pi} {4}-i \sin\frac {\pi} {4} \biggl]$$= \bar{z}$

$z^1= r[\cos(-\theta)+i \sin(-\theta)$

$= \bar{z}$

$(r,\theta + 2k{\pi}), \ k\in \Z \mapsto x= r\cos\theta, y=r\sin\theta$

$\theta_\circ \leq\theta \leq \theta_\circ +2 {\pi}$

$\theta_\circ =0 \quad 0\leq \theta <2 {\pi}$

${\pi} < \theta \leq {\pi}$

Principal argument of $z= \theta \in (-\pi,\ \pi]$

(3) $r=0 \quad \theta \in[0,2\pi]$

$(0,\theta)$ represents the origin.

It means that we do not have well defined polar representation for the origin.

Observation $\ \ \tan(x+\pi) = \tan x$
$\tan(x+k\pi) = \tan x, \$ for $k\in \Z$
Find $\theta \in (-\pi,\pi)$ such that $\ \ \tan(\theta) =\frac {y} {x}$
$(\theta) = \tan^{-1}\biggl(\frac {y} {x}\biggl) + k\pi, \quad k\in \Z$

Restrict $\tan^{-1}\biggl(\frac {y} {x}\biggl)$ in the interval $(\frac {-\pi} {2}, \frac {\pi} {2})$

which we call it as arc $\tan\frac {y} {x}$

$\theta =$ arc $\tan\frac {y} {x}+k_+\pi$

If $x \neq 0$

arg (z) = arc $\tan \left(\frac{y}{x}\right)+k_{+} \pi$

Where $k_{+} =$

$z=-1 + i$

$r= \sqrt{1 +1} =\sqrt2$

$\theta=$ arc $\tan\frac{1}{-1} +k_+\pi$

$\theta = \frac {-\pi} {4} +\pi$

$\theta = \frac {3} {4}\pi$

Exercise: $Z= \sqrt{2} + 2 \sqrt{3}i$

$Z= 1+\cos a +i \sin a,a\in (0,2\pi)$

$r= \sqrt{(1+\cos c)^2 + sin^2 c} = \sqrt {2+2\cos a}$

$r= \sqrt {2(1+\cos a)} = \sqrt {2.2 cos^2 \frac {a} {2}}$

$r= 2|\cos \frac {a} {2}|$

For $a\in(0,\pi),\quad z$ is in the $1^{st}$ quadrant

$\theta =$ arc $\tan \biggl(\frac {\sin a} {1+\cos a}\biggl) =$ arc $\tan \biggl(\frac {2\sin\frac {a} {2} \cos \frac {a}{2}} {2\cos^2\frac{a} {2}}\biggl)$

arc $\tan (\tan (\frac {a} {2})) = \frac {a} {2}$

$a\in (\pi, 2\pi)$

Verify that $\theta =\frac {a} {2} -\pi$

$a=\pi, \quad z= 1+\cos\pi+i\sin\pi$

$= 0$

$z_1=r_1$ cis $\theta_1 \quad z_2 = r_2$ cis $\theta_2$

$z_1. z_2 = r_1r_2[\cos\theta_1 -\sin\theta_1, \sin\theta_2+i(\cos\theta_1\sin\theta_2+\sin\theta_1\cos\theta_2)]$

$z_1. z_2 =r_1r_2$ cis $(\theta_1+\theta_2)$

Remark: $|z_1. z_2| =r_1r_2 =|z_1| |z_2|$

arg $(z_1 z_2) =$ arg $(z_1) +$ arg $(z_2) + k_+2\pi$

where $$k_{+}=\{\begin{array}{ll}0-\pi <arg(z_1)+arg(z_2)\le \pi & \\ 1arg(z_1)+arg(z_2)\le -\pi \\ -1arg(z_1)+arg(z_2)>\pi \end{array}$$

$z_1=1+i \quad z_2 = \sqrt{3} +i$

$z_1=\sqrt{2}$ cis $\biggl(\frac {\pi} {4}\biggl), z_2 = 2$ cis $\biggl(\frac {\pi} {6}\biggl)$

$z_1 z_2 = 2\sqrt{2}$ cis $\biggl(\frac {5\pi} {12}\biggl)$

De Moivre’s Formula

$z=r(\cos\theta + i\sin\theta)$

$z^n=r^n(\cos n\theta + i \sin n\theta)\quad n\geq1$

$z^2=z.z =r^2(\cos2\theta + i\sin 2\theta)$

$z_1=1+i$

Calculate $z^{1000}$

$(1+i)^{1000} = [\sqrt{2}$ cis $(\frac {\pi} {4})]^{1000}$

$(1+i)^{1000}=2^{500}$ cis $(250\pi)$

$(1+i)^{1000}=2^{500}$

Exercise: prove that

$\sin 5\theta=16 \sin^5\theta-20 \sin^3\theta +5\sin\theta$

$\cos 5\theta=16 \cos^5\theta-20 \cos^3\theta +5\cos\theta$

Hint: cis $5\theta=($cis $\theta)^5$

Thank you