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How does plotting a complex number reveal its size and direction through modulus and argument?

Represent complex numbers on the Argand plane and find modulus and argument, converting to polar form

WACE Specialist Unit 3 complex plane: plotting on the Argand diagram, the modulus as distance from the origin, the argument and principal argument in the interval negative pi to pi, quadrant checks, and conversion to polar form, with a worked example.

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The Argand plane
  3. Modulus
  4. Argument
  5. Conversion to polar form
  6. Quadrant-by-quadrant argument rules
  7. Modulus as a distance and the triangle inequality

What this dot point is asking

SCSA wants you to plot complex numbers, compute modulus and argument correctly with attention to quadrant, and convert between Cartesian and polar form.

The Argand plane

The Argand plane represents z=x+iyz = x + iy as the point (x,y)(x, y), with the horizontal axis the real axis and the vertical axis the imaginary axis. Addition corresponds to vector addition and conjugation corresponds to reflection in the real axis. This geometric picture is what makes polar form and loci natural.

Modulus

The modulus is multiplicative: zw=zw|zw| = |z|\,|w| and zw=zw\left|\tfrac{z}{w}\right| = \tfrac{|z|}{|w|}. It also satisfies z2=zzˉ|z|^2 = z\bar z and the triangle inequality z+wz+w|z + w| \le |z| + |w|.

Argument

The argument of zz is the angle θ\theta, measured anticlockwise from the positive real axis, such that x=zcosθx = |z|\cos\theta and y=zsinθy = |z|\sin\theta. Because adding 2π2\pi returns the same point, the argument is only defined up to multiples of 2π2\pi. The principal argument Argz\operatorname{Arg} z is the unique value in (π,π](-\pi, \pi].

To find it, compute the reference angle α=tan1 ⁣yx\alpha = \tan^{-1}\!\left|\tfrac{y}{x}\right| and then place it in the correct quadrant by inspecting the signs of xx and yy. A sketch settles the sign every time.

Conversion to polar form

Once you have r=zr = |z| and θ=argz\theta = \arg z, the polar (modulus-argument) form is

z=r(cosθ+isinθ)=rcisθ.z = r(\cos\theta + i\sin\theta) = r\,\text{cis}\,\theta.

Going the other way, x=rcosθx = r\cos\theta and y=rsinθy = r\sin\theta recovers Cartesian form.

Quadrant-by-quadrant argument rules

It pays to know the adjustment in each quadrant once you have the reference angle α=tan1 ⁣yx\alpha = \tan^{-1}\!\left|\dfrac{y}{x}\right|. In the first quadrant (x>0x > 0, y>0y > 0) the argument is α\alpha. In the second quadrant (x<0x < 0, y>0y > 0) it is πα\pi - \alpha. In the third quadrant (x<0x < 0, y<0y < 0) it is (πα)-(\pi - \alpha), i.e. απ\alpha - \pi. In the fourth quadrant (x>0x > 0, y<0y < 0) it is α-\alpha. On the axes, read the angle directly: arg(positive real)=0\arg(\text{positive real}) = 0, arg(positive imaginary)=π2\arg(\text{positive imaginary}) = \dfrac{\pi}{2}, arg(negative real)=π\arg(\text{negative real}) = \pi, and arg(negative imaginary)=π2\arg(\text{negative imaginary}) = -\dfrac{\pi}{2}.

Modulus as a distance and the triangle inequality

Because zw|z - w| is the distance between the points zz and ww in the Argand plane, the modulus is the bridge between algebra and geometry that powers the loci topic. The triangle inequality z+wz+w|z + w| \le |z| + |w| is the statement that one side of a triangle cannot exceed the sum of the other two, with equality only when zz and ww point the same way. The reverse triangle inequality zwzw\big||z| - |w|\big| \le |z - w| is also worth knowing for bounding arguments. These distance interpretations recur whenever a question mixes modulus conditions with geometry.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20235 marksCalculator-free. (a) Find the modulus and principal argument of z=22iz = -2 - 2i. (b) Hence write zz in polar form, with the argument in (π,π](-\pi, \pi].
Show worked answer →

A modulus-and-argument conversion.

(a) Modulus: z=(2)2+(2)2=8=22|z| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}. The point (2,2)(-2, -2) is in the third quadrant (x<0x < 0, y<0y < 0). The reference angle is tan1 ⁣(22)=π4\tan^{-1}\!\left(\dfrac{2}{2}\right) = \dfrac{\pi}{4}. In the third quadrant the principal argument is (ππ4)=3π4-\left(\pi - \dfrac{\pi}{4}\right) = -\dfrac{3\pi}{4}.

(b) So z=22cis ⁣(3π4)z = 2\sqrt{2}\,\text{cis}\!\left(-\dfrac{3\pi}{4}\right), which lies in (π,π](-\pi, \pi].

Markers reward the modulus 222\sqrt{2}, the third-quadrant placement, and the principal argument 3π4-\dfrac{3\pi}{4}.

WACE 20204 marksCalculator-assumed. Given z=5|z| = 5 and argz=2π3\arg z = \dfrac{2\pi}{3}, find zz in Cartesian form, and state z2|z^2| and arg(z2)\arg(z^2).
Show worked answer →

Conversion plus the multiplicative properties.

Cartesian: z=5cos2π3+5isin2π3=5(12)+5i(32)=52+532iz = 5\cos\dfrac{2\pi}{3} + 5i\sin\dfrac{2\pi}{3} = 5\left(-\dfrac{1}{2}\right) + 5i\left(\dfrac{\sqrt{3}}{2}\right) = -\dfrac{5}{2} + \dfrac{5\sqrt{3}}{2}i.

For z2z^2: the modulus is multiplicative, so z2=z2=25|z^2| = |z|^2 = 25, and the argument adds, so arg(z2)=2argz=4π3\arg(z^2) = 2\arg z = \dfrac{4\pi}{3}, which as a principal value is 2π3-\dfrac{2\pi}{3}.

Markers reward the Cartesian form, z2=25|z^2| = 25, and arg(z2)=4π3\arg(z^2) = \dfrac{4\pi}{3} or equivalently 2π3-\dfrac{2\pi}{3}.

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