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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=[    zmodulus,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=zcos(θ)+izsin(θ)z=z(cos(θ)+isin(θ))z=zcis(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
360360^\circ2π2\pi
270270^\circ32π\frac{3}{2}\pi
180180^\circπ\pi
135135^\circ34π\frac{3}{4}\pi
9090^\circ12π\frac{1}{2}\pi
6060^\circ13π\frac{1}{3}\pi
4545^\circ14π\frac{1}{4}\pi
3030^\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+yiz=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=tan1yx\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=πα\theta = \pi - \alphaπ2<θ<π\frac{\pi}{2} < \theta < \pi
IIILower Leftx<0,y<0x < 0, y < 0θ=(πα)\theta = -(\pi - \alpha)π<θ<π2-\pi < \theta < -\frac{\pi}{2}
IVLower Rightx>0,y<0x > 0, y < 0θ=α\theta = -\alphaπ2<θ<0-\frac{\pi}{2} < \theta < 0