The Complex Plane and Coordinate Geometry

Problem Statement

In standard mathematics, squaring a real number always produces a positive result. This meant equations like $x^2 + 1 = 0$ had no solution. Mathematicians solved this by introducing the imaginary unit $i$, defined by $i^2 = -1$.

A complex number $z = x + iy$ consists of a real part ($x$) and an imaginary part ($y$). To visualise these numbers, we cannot use a standard 1D number line. We must use a 2D plane called the Argand Diagram (or Complex Plane). The horizontal axis represents the real part ($\text{Re}$), and the vertical axis represents the imaginary part ($\text{Im}$).

Plotting complex numbers on the complex plane relates perfectly to plotting points on a Cartesian plane. Consequently, standard coordinate geometry methods—like finding the distance between two points—apply directly to complex numbers via the Modulus function $|z|$.

Consider three complex numbers:

• $z_1 = 3 + 4i$
• $z_2 = -2 + i$
• $z_3 = 3 - 4i$

(a) Plot these three complex numbers on an Argand diagram. (Describe their positions).

(b) The modulus $|z|$ represents the distance of $z$ from the origin. Calculate the exact values of $|z_1|$ and $|z_2|$. What does the result for $z_1$ and $z_3$ tell you about complex conjugates?

(c) The distance between two complex numbers $A(z_a)$ and $B(z_b)$ is given by $|z_a - z_b|$. Find the exact distance between $z_1$ and $z_2$.

Hints

• Part (a): Treat $z = x + iy$ just like the Cartesian coordinate $(x, y)$.
• Part (b): The formula for the modulus is $|x + iy| = \sqrt{x^2 + y^2}$. This is simply the Pythagorean theorem measuring the hypotenuse from the origin. Notice that $z_3$ is the conjugate of $z_1$ (denoted as $\bar{z}_1$).
• Part (c): First, subtract the two complex numbers algebraically by grouping the real parts and imaginary parts together: $(x_1 - x_2) + i(y_1 - y_2)$. Then, find the modulus of this new complex number.

Solutions

Part (a): Plotting the Points

1. To plot $z_1 = 3 + 4i$, move 3 units right along the Real axis, and 4 units up along the Imaginary axis. This corresponds to the point $(3, 4)$.
2. To plot $z_2 = -2 + i$, move 2 units left on the Real axis, and 1 unit up on the Imaginary axis. This corresponds to the point $(-2, 1)$.
3. To plot $z_3 = 3 - 4i$, move 3 units right on the Real axis, and 4 units down on the Imaginary axis. This corresponds to the point $(3, -4)$.

Part (b): Calculating the Modulus

1. Calculate $|z_1|$: $|z_1| = |3 + 4i| = \sqrt{3^2 + 4^2}$ $|z_1| = \sqrt{9 + 16} = \sqrt{25} = 5$ (The point $z_1$ is 5 units away from the origin).
2. Calculate $|z_2|$: $|z_2| = |-2 + i| = \sqrt{(-2)^2 + 1^2}$ $|z_2| = \sqrt{4 + 1} = \sqrt{5}$
3. Calculate $|z_3|$: $|z_3| = |3 - 4i| = \sqrt{3^2 + (-4)^2}$ $|z_3| = \sqrt{9 + 16} = \sqrt{25} = 5$
4. Conclusion: We see that $|z_1| = |z_3|$. Since $z_3$ is the complex conjugate of $z_1$ (the imaginary sign is flipped), we can conclude a fundamental rule: A complex number and its conjugate always have the exact same modulus ($|z| = |\bar{z}|$). Geometrically, they are reflections of each other across the real axis.

Part (c): Distance Between Two Points

1. We need to find the distance $|z_1 - z_2|$.
2. First, calculate the complex difference: $z_1 - z_2 = (3 + 4i) - (-2 + i)$ $z_1 - z_2 = (3 - (-2)) + (4i - i)$ $z_1 - z_2 = 5 + 3i$
3. Now, find the modulus of this new complex number: $|z_1 - z_2| = |5 + 3i| = \sqrt{5^2 + 3^2}$ $|z_1 - z_2| = \sqrt{25 + 9}$ $|z_1 - z_2| = \sqrt{34}$
4. The exact distance between the two points on the complex plane is $\sqrt{34}$ units.

Takeaways

• The 2D Analogy: Always think of a complex number $x + iy$ as a point $(x, y)$ or a vector $\begin{pmatrix} x \\ y \end{pmatrix}$. This makes visualising problems much easier.
• The Modulus is a Distance: The modulus function is simply a generalisation of the absolute value function on a number line, extended to a 2D metric space. It always returns a positive real number representing physical length.
• Conjugate Reflections: The complex conjugate $\bar{z}$ is geometrically just a reflection of $z$ across the $x$-axis (the Real axis). This is why their distances to the origin are identical.

Further Readings

• HSC Complex Numbers: https://vumaths.com/booklets/hsc-complex-numbers/
• HSC Vectors: https://vumaths.com/booklets/hsc-vectors/
• HSC Collections: https://vumaths.com/booklets/hsc-collections/

Connect with me

If you found graphing complex numbers intuitive, there is an entire world of complex geometry waiting for you! Check out my complete booklets on Vu's Maths Hub. I break down everything from locus problems to roots of unity. Join my YouTube channel for visual walkthroughs, follow my Instagram for daily maths tips, and subscribe to my Substack to learn how complex numbers drive modern physics!