Argand diagrams, modulus and argument | CAIE 9709 Pure Mathematics 3 Notes

Chapters

Pure Mathematics 3

Mathematics9709

Modulus is length; argument is direction

z = −3 + 2i lies in quadrant II

ReIm

−3 + 2i

modulus = distance = √13

argument = π − tan⁻¹(2/3)

Modulus is distance; argument is a directed angle from the positive real axis.

Use equal scales on the real and imaginary axes. Pythagoras gives the distance from the origin:

|z|=\sqrt{x^2+y^2}

\arg z=\theta

The principal argument usually lies in $(-\pi,\pi]$ .
The syllabus also permits $[0,2\pi)$ unless the question specifies an interval.
Zero has modulus 0, but its argument is undefined.

tan⁻¹(y/x) needs a quadrant check

A calculator value from $\tan^{-1}(y/x)$ gives a reference angle. Sketch the point first, then place that angle in the correct quadrant. An atan2 calculation performs this adjustment directly.

Worked example

Find modulus and principal argument

Let $z=-3+2i$ .

|z|=\sqrt{(-3)^2+2^2}

=\sqrt{13}

The point is in quadrant II, so add the quadrant information rather than accepting a negative calculator output.

\arg z=\pi-\tan^{-1}\left(\frac23\right)

Answer

|z|=\sqrt{13}

\arg z\approx2.55\text{ rad}

Common mistake

\tan^{-1}(y/x)

does not identify the quadrant. Sketch the point or use an atan2 calculation, then express the angle in the interval requested.

Move between Cartesian and polar descriptions

x=r\cos\theta,\qquad y=r\sin\theta

z=r(\cos\theta+i\sin\theta)=re^{i\theta}

The first expression is trigonometric form. The shorter expression $re^{i\theta}$ is exponential form; here it records the same modulus r and argument θ.

Your turnIndependent Argand check [4 marks]

Express $z=-1-\sqrt3i$ in the form $re^{i\theta}$ , using $-\pi<\theta\le\pi$ .

Show worked answer

The point is in quadrant III, represented by a negative principal angle.

Answer

z=2e^{-2\pi i/3}

(Pure Mathematics 2 & 3 Coursebook, Ch. 11.3)

Modulus is length; argument is direction

z = −3 + 2i lies in quadrant II

ReIm

−3 + 2i

modulus = distance = √13

argument = π − tan⁻¹(2/3)

Modulus is distance; argument is a directed angle from the positive real axis.

Use equal scales on the real and imaginary axes. Pythagoras gives the distance from the origin:

|z|=\sqrt{x^2+y^2}

\arg z=\theta

The principal argument usually lies in $(-\pi,\pi]$ .
The syllabus also permits $[0,2\pi)$ unless the question specifies an interval.
Zero has modulus 0, but its argument is undefined.

tan⁻¹(y/x) needs a quadrant check

A calculator value from $\tan^{-1}(y/x)$ gives a reference angle. Sketch the point first, then place that angle in the correct quadrant. An atan2 calculation performs this adjustment directly.

Worked example

Find modulus and principal argument

Let $z=-3+2i$ .

|z|=\sqrt{(-3)^2+2^2}

=\sqrt{13}

The point is in quadrant II, so add the quadrant information rather than accepting a negative calculator output.

\arg z=\pi-\tan^{-1}\left(\frac23\right)

Answer

|z|=\sqrt{13}

\arg z\approx2.55\text{ rad}

Common mistake

\tan^{-1}(y/x)

does not identify the quadrant. Sketch the point or use an atan2 calculation, then express the angle in the interval requested.

Move between Cartesian and polar descriptions

x=r\cos\theta,\qquad y=r\sin\theta

z=r(\cos\theta+i\sin\theta)=re^{i\theta}

The first expression is trigonometric form. The shorter expression $re^{i\theta}$ is exponential form; here it records the same modulus r and argument θ.

Your turnIndependent Argand check [4 marks]

Express $z=-1-\sqrt3i$ in the form $re^{i\theta}$ , using $-\pi<\theta\le\pi$ .

Show worked answer

The point is in quadrant III, represented by a negative principal angle.

Answer

z=2e^{-2\pi i/3}

(Pure Mathematics 2 & 3 Coursebook, Ch. 11.3)