Category Archives: Sketching Complex Regions

February 25, 2026 · 1:22 pm

Complex Locus Question

My Year 12 Specialist students are working on complex loci again. The following type of question always creates confusion.

z \in \mathbb{C}, w_1=1+i

w_2=1-i

$arg(z-w_1)$ is the angle the vector from $w_1$ to $z$ makes with the positive $x-$ axis, likewise for $arg(z-w_2)$ .

I am going to plot a possible $z$ and try to see the geometry that works.

We want $arg(z-w_1)-arg(z-w_2)=\frac{\pi}{6}$

I am going to take advantage of some triangle geometry

Using the External Angle Theorem, we know $\alpha=\beta+\theta$

$\begin{equation*}arg(z-w_1)-arg(z-w_2)=(\alpha+\frac{\pi}{2})-(\beta+\frac{\pi}{2}0=\alpha-\beta=\theta\end{equation}$

Therefore $\theta=\frac{\pi}{6}$

So we want all of the $z$ values that have an angle of $\frac{\pi}{6}$

Now we are going to use some circle geometry -The angle at the circumference subtended by the same arc are congruent. So we need to find a circle that has those three points ( $z, w_1$ and $w_2$ ) on the circumference.

Hence the locus is

Now we need to find the radius and centre of the circle.

Using another circle theorem, the angle at the centre is twice the angle at the circumference.

The triangle must be equilateral (it is isosceles with a vertex angle of $\frac{\pi}{3}$ )

Hence the radius is 2.

$h=\sqrt{2^2-1^2}=\sqrt{3}$

Hence the centre is $(-\sqrt{3}+1, 0)$

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Filed under Circle Theorems, Complex Numbers, Geometry, Interesting Mathematics, Pythagoras, Sketching Complex Regions, Year 12 Specialist Mathematics

Tagged as complex loci

January 22, 2026 · 9:27 am

Complex Loci Question

A sketch of the locus of a complex number $z$ is shown above, determine the maximum value of $arg(z)$ correct to two decimal places where $0\le z \le 2\pi$

Draw tangent lines from the origin to the circle.

Remember tangent lines are perpendicular to the radii

The maximum argument is this angle

I am going to find the angle in two sections

From the diagram the radius of the circle is $2$ and the centre is $(4, 3)$ . Hence the distance from the origin to the centre is $5$ .

$\begin{equation*}sin(\theta_1)=\frac{2}{5}\end{equation}$

$\begin{equation*}\theta_1=0.412\end{equation}$

$\begin{equation*}sin(\theta_2)=\frac{3}{5}\end{equation}$

$\begin{equation*}\theta_2=0.644\end{equation}$

Hence maximum $arg(z)=1.06$

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Filed under Complex Numbers, Right Trigonometry, Sketching Complex Regions, Trigonometry, Year 12 Specialist Mathematics

Tagged as argument of a complex number, complex locus, right trig

August 23, 2024 · 7:41 am

Sketching in the Complex Plane Using a TI-nspire CX 11

Let $R$ be the region of the complex plane where the inequalities $|z-i|\le2$ and $|z-\bar{z}|\ge3$ hold simultaneously.

First find the Cartesian equations.

Second, sketch each of the functions.

The section that is shaded twice is our region.

Determine the minimum value of $Re(z)$ in $R$ .

We can find the point of intersection between the circle and the line.

$Re(z)=-1.94$

Or if you want exact values

Use the Solve Systems of Equations tool – Menu – Algebra – Solve Systems of Equations – Solve Systems of Equations.

$Re(z)=-\frac{\sqrt{15}}{2}$

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Filed under Complex Numbers, Sketching Complex Regions, TI nspire CX 11

April 17, 2024 · 9:32 am

Sketching Subsets of the Complex Plane – Problem 2

I found this question on the madasmaths site – his resources are fabulous.

I am not sure I would have been able to do this in an exam.

I have split my solution into 6 images and there is a pdf version at the bottom

PDF version of my solution

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Filed under Complex Numbers, Sketching Complex Regions

Tagged as complex loci, complex numbers, sketching complex regions