Finding complex modulus
∣ z ∣ = r = x 2 + y 2 {|z| = r =\sqrt{x^2+y^2}} ∣ z ∣ = r = x 2 + y 2 Finding complex argument
Let arg z = θ \arg{z} = \theta arg z = θ tan θ = y x {\tan \theta = \frac{y}{x}} tan θ = x y α = tan − 1 ∣ y x ∣ {\alpha = \tan^{-1} \left| \frac{y}{x} \right|} α = tan − 1 ∣ ∣ x y ∣ ∣
θ = { α if x > 0 , y > 0 π − α if x < 0 , y > 0 − ( π − α ) if x < 0 , y < 0 − α if x > 0 , y < 0 \theta = \begin{cases}
\alpha & \text{if } x > 0, y > 0 \\
\pi - \alpha & \text{if } x < 0, y > 0 \\
-(\pi - \alpha) & \text{if } x < 0, y < 0 \\
-\alpha & \text{if } x > 0, y < 0
\end{cases} θ = ⎩ ⎨ ⎧ α π − α − ( π − α ) − α if x > 0 , y > 0 if x < 0 , y > 0 if x < 0 , y < 0 if x > 0 , y < 0 The trigonometric and exponential forms
After finding r r r θ , \theta, θ ,
Cartesian form z = x + y i Polar/trigonometric z = r ( cos θ + i sin θ ) Exponential form z = r e i θ \begin{gather*}
\textbf{Cartesian form } \\
z = x+y\mathrm{i} \\
\textbf{Polar/trigonometric } \\
z = r(\cos \theta + \mathrm{i}\sin \theta) \\
\textbf{Exponential form } \\
z = r\mathrm{e}^{\mathrm{i}\theta}
\end{gather*} Cartesian form z = x + y i Polar/trigonometric z = r ( cos θ + i sin θ ) Exponential form z = r e i θ Examples
Q1
Q2
Q3
Q4
Real
Purely Imaginary
Random z = − 1 + 3 i z = - 1 + \sqrt{3} \mathrm{i} z = − 1 + 3 i ∣ z ∣ = r = x 2 + y 2 = ( − 1 ) 2 + ( 3 ) 2 = 2 \begin{aligned}
& |z| = r \\
&= \sqrt{x^2 + y^2} \\
&= \sqrt{(- 1)^2 + (\sqrt{3})^2} \\
&= 2
\end{aligned} ∣ z ∣ = r = x 2 + y 2 = ( − 1 ) 2 + ( 3 ) 2 = 2 α = tan − 1 ∣ y x ∣ = tan − 1 3 1 = 1 3 π \begin{aligned}
\alpha &= \tan^{-1}\left|\frac{y}{x}\right| \\
&= \tan^{-1} \frac{\sqrt{3}}{1} \\
&= \frac{1}{3} \pi
\end{aligned} α = tan − 1 ∣ ∣ x y ∣ ∣ = tan − 1 1 3 = 3 1 π arg ( z ) = θ = π − α = 2 3 π \begin{aligned}
& \arg (z) = \theta \\
& = \pi - \alpha \\
& = \frac{2}{3} \pi
\end{aligned} arg ( z ) = θ = π − α = 3 2 π r = 2 , θ = 2 3 π \boxed{r=2, \theta = \frac{2}{3} \pi} r = 2 , θ = 3 2 π Polar/trigo form: z = r ( cos θ + i sin θ ) z = 2 ( cos 2 3 π + i sin 2 3 π ) \begin{aligned}
& \textrm{Polar/trigo form:} \\
& z = r(\cos \theta + \mathrm{i} \sin \theta) \\
& \phantom{z } = 2 \left( \cos \frac{2}{3} \pi + \mathrm{i} \sin \frac{2}{3} \pi \right)
\end{aligned} Polar/trigo form: z = r ( cos θ + i sin θ ) z = 2 ( cos 3 2 π + i sin 3 2 π ) Exponential form: z = r e i θ z = 2 e 2 3 π i \begin{aligned}
& \textrm{Exponential form:} \\
& z = r\mathrm{e}^{\mathrm{i}\theta} \\
& \phantom{z } = 2 \mathrm{e}^{ \frac{2}{3} \pi \mathrm{i} }
\end{aligned} Exponential form: z = r e i θ z = 2 e 3 2 π i Conversion from polar form to cartesian form
The trigonometric form is the key to conversion back to cartesian form. Given
z = r e i θ = r ( cos θ + i sin θ ) , z = r\mathrm{e}^{\mathrm{i}\theta} \allowbreak {= r(\cos\theta + \mathrm{i}\sin\theta),} z = r e i θ = r ( cos θ + i sin θ ) , r cos θ r\cos\theta r cos θ r sin θ r\sin\theta r sin θ
Examples
Convert 2 e 1 6 π i {\displaystyle 2 \mathrm{e}^{ \frac{1}{6} \pi \mathrm{i} }} 2 e 6 1 π i x + y i {x+y\mathrm{i}} x + y i
z = 2 e 1 6 π i = 2 ( cos 1 6 π + i sin 1 6 π )
\begin{aligned}
z &= 2 \mathrm{e}^{ \frac{1}{6} \pi \mathrm{i} } \\
&= 2 \left( \cos \frac{1}{6} \pi + \mathrm{i} \sin \frac{1}{6} \pi \right)
\end{aligned}
z = 2 e 6 1 π i = 2 ( cos 6 1 π + i sin 6 1 π )