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Modulus and Argument of Complex Numbers

How to write complex numbers in modulus-argument form

created December 8, 2025 updated May 31, 2026 2 min read

COMPLEX NUMBER REPRESENTATION

An imaginary (complex) number zz can be written as:

z=x+yiz=[    ∣z∣↑modulus,arg(z)↑argument    ]\begin{aligned} z &= x + yi \\ z &= [ \;\; \underset{\begin{subarray}{c} \uparrow \\ \text{modulus} \end{subarray}}{|z|} , \quad \underset{\begin{subarray}{c} \uparrow \\ \text{argument} \end{subarray}}{\text{arg}(z)} \;\; ] \end{aligned}

Other forms

z=x+iyz = x + iy can be rewritten as:

z=∣z∣cos⁡(θ)+i∣z∣sin⁡(θ)z=∣z∣(cos⁡(θ)+isin⁡(θ))∴z=∣z∣cis⁡(θ)where cis⁡(θ)=cos⁡(θ)+isin⁡(θ)\begin{aligned} z &= |z|\cos(\theta) + i|z|\sin(\theta) \\ z &= |z|(\cos(\theta) + i\sin(\theta)) \\ \therefore z &= |z|\operatorname{cis}(\theta) && \text{where } \operatorname{cis}(\theta) = \cos(\theta) + i\sin(\theta) \end{aligned}

Degrees and Radians

To switch between degrees and radians, use these constants:

1∘=π1801 rad=180π\boxed{1^\circ = \frac{\pi}{180}} \quad \boxed{1 \text{ rad} = \frac{180}{\pi}}
DegreesRadians
360∘360^\circ2π2\pi
270∘270^\circ32π\frac{3}{2}\pi
180∘180^\circπ\pi
135∘135^\circ34π\frac{3}{4}\pi
90∘90^\circ12π\frac{1}{2}\pi
60∘60^\circ13π\frac{1}{3}\pi
45∘45^\circ14π\frac{1}{4}\pi
30∘30^\circ16π\frac{1}{6}\pi

Finding the Modulus and Argument

Finding the modulus of a complex number is very simple, we just use Pythagoreas' Theorem.

z=x+yi∣z∣=x2+y2\begin{aligned} z &= x + yi \\ |z| &= \sqrt{x^2+y^2} \end{aligned}

The argument of a complex number z=x+iyz = x + iy is the angle θ\theta made with the positive real axis (cis(θ)\text{cis}(\theta)). Because tan⁡(θ)=yx\tan(\theta) = \frac{y}{x}, we use the signs of xx and yy to determine the correct quadrant.

First, find the reference angle: α=tan⁡−1∣yx∣\alpha = \tan^{-1} \left| \frac{y}{x} \right|.

QuadrantLocationSignsCalculationRange of θ\theta
IUpper Rightx>0,y>0x > 0, y > 0θ=α\theta = \alpha0<θ<π20 < \theta < \frac{\pi}{2}
IIUpper Leftx<0,y>0x < 0, y > 0θ=π−α\theta = \pi - \alphaπ2<θ<π\frac{\pi}{2} < \theta < \pi
IIILower Leftx<0,y<0x < 0, y < 0θ=−(π−α)\theta = -(\pi - \alpha)−π<θ<−π2-\pi < \theta < -\frac{\pi}{2}
IVLower Rightx>0,y<0x > 0, y < 0θ=−α\theta = -\alpha−π2<θ<0-\frac{\pi}{2} < \theta < 0