Complex Numbers and Quadratic Equations • Topic 2 of 3

Modulus, Conjugate and the Argand Plane

Every complex number $z = a + ib$ can be pictured as the point $(a, b)$ in a plane: the horizontal axis carries the real part and the vertical axis carries the imaginary part. This picture is the Argand plane (or complex plane), and it turns algebra into geometry.

Conjugate. The conjugate of $z = a + ib$ is obtained by flipping the sign of the imaginary part:

$$\overline{z} = a - ib$$

Geometrically, $\overline{z}$ is the mirror image of $z$ across the real axis. A key product collapses to a real number:

$$z\,\overline{z} = (a + ib)(a - ib) = a^2 + b^2$$

Modulus. The modulus $|z|$ is the distance of the point from the origin, found by Pythagoras:

$$|z| = \sqrt{a^2 + b^2}, \qquad z\,\overline{z} = |z|^2$$

Because $z\,\overline{z} = |z|^2$, the multiplicative inverse of a non-zero $z$ has a clean closed form — there is no need to guess:

$$z^{-1} = \dfrac{\overline{z}}{|z|^2} = \dfrac{a - ib}{a^2 + b^2}$$

Polar form. Instead of the coordinates $(a, b)$, a point can be located by its distance $r = |z|$ from the origin and the angle $\theta$ its line makes with the positive real axis. Then $a = r\cos\theta$ and $b = r\sin\theta$, which gives the polar (or trigonometric) form:

$$z = r(\cos\theta + i\sin\theta), \qquad r = \sqrt{a^2 + b^2}, \quad \tan\theta = \dfrac{b}{a}$$

The angle $\theta$ is the argument of $z$. The value taken in $(-\pi, \pi]$ is the principal argument, written $\arg(z)$. The quadrant of the point decides the correct angle, so you must check signs of $a$ and $b$ rather than trust $\tan^{-1}(b/a)$ blindly.

Quadrant of $(a,b)$SignsPrincipal argument $\theta$
I$a > 0,\ b > 0$$\tan^{-1}\dfrac{b}{a}$
II$a < 0,\ b > 0$$\pi - \tan^{-1}\dfrac{b}{|a|}$
III$a < 0,\ b < 0$$-\pi + \tan^{-1}\dfrac{|b|}{|a|}$
IV$a > 0,\ b < 0$$-\tan^{-1}\dfrac{|b|}{a}$

Deeper Insight — the modulus and conjugate are two halves of one idea: The conjugate and the modulus are not separate tricks; they are bound together by the single identity $z\,\overline{z} = |z|^2$. That identity is exactly why division and inversion of complex numbers work at all: multiplying by the conjugate is the move that converts a complex denominator into a real one, and the real number it produces is precisely $|z|^2$. Geometry makes the rest intuitive — $|z|$ is a length, so $|z_1 z_2| = |z_1||z_2|$ (lengths multiply) and the modulus obeys the triangle inequality $|z_1 + z_2| \le |z_1| + |z_2|$ just as distances do. The polar form then reveals the deepest fact of all: multiplying complex numbers adds their arguments and multiplies their moduli, so multiplication by a complex number is a rotation combined with a scaling. Seeing complex numbers as points with a length and a direction, rather than as $a + ib$ alone, is what makes the next chapters on rotations and roots fall into place.

Point z on the Argand plane showing modulus and argument The Argand Plane ReIm z = a + ib ab r = |z| θ Conjugate as a reflection across the real axis Conjugate: Mirror Across the Real Axis ReIm z = a + ib z̅ = a − ib
1
Worked Example
Find the conjugate and modulus of $z = 4 - 3i$.
Solution
  1. Conjugate: flip the sign of the imaginary part, $\overline{z} = 4 + 3i$.
  2. Modulus: $|z| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25}$.

Answer: $\overline{z} = 4 + 3i,\ |z| = 5$

2
Worked Example
Find the multiplicative inverse of $z = 2 + 5i$.
Solution
  1. Use $z^{-1} = \dfrac{\overline{z}}{|z|^2}$ with $\overline{z} = 2 - 5i$.
  2. $|z|^2 = 2^2 + 5^2 = 4 + 25 = 29$.
  3. So $z^{-1} = \dfrac{2 - 5i}{29}$.

Answer: $z^{-1} = \dfrac{2}{29} - \dfrac{5}{29}i$

3
Worked Example
Express $z = 1 + i$ in polar form.
Solution
  1. Modulus: $r = \sqrt{1^2 + 1^2} = \sqrt{2}$.
  2. The point $(1, 1)$ lies in the first quadrant, so $\theta = \tan^{-1}\dfrac{1}{1} = \dfrac{\pi}{4}$.
  3. Substitute into $z = r(\cos\theta + i\sin\theta)$.

Answer: $z = \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right)$

4
Worked Example
Find the modulus and principal argument of $z = -1 + \sqrt{3}\,i$.
Solution
  1. Modulus: $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$.
  2. The point $(-1, \sqrt{3})$ is in the second quadrant, so use $\theta = \pi - \tan^{-1}\dfrac{\sqrt{3}}{1}$.
  3. $\tan^{-1}\sqrt{3} = \dfrac{\pi}{3}$, so $\theta = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$.

Answer: $r = 2,\ \arg(z) = \dfrac{2\pi}{3}$

5
Worked Example
Write $z = -\sqrt{3} - i$ in polar form using its principal argument.
Solution
  1. Modulus: $r = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$.
  2. The point $(-\sqrt{3}, -1)$ is in the third quadrant, so $\theta = -\pi + \tan^{-1}\dfrac{1}{\sqrt{3}}$.
  3. $\tan^{-1}\dfrac{1}{\sqrt{3}} = \dfrac{\pi}{6}$, giving $\theta = -\pi + \dfrac{\pi}{6} = -\dfrac{5\pi}{6}$.

Answer: $z = 2\left(\cos\left(-\dfrac{5\pi}{6}\right) + i\sin\left(-\dfrac{5\pi}{6}\right)\right)$

6
Worked Example
If $z = 3 - 4i$, verify that $z\,\overline{z} = |z|^2$.
Solution
  1. $\overline{z} = 3 + 4i$, so $z\,\overline{z} = (3 - 4i)(3 + 4i) = 9 - 16i^2 = 9 + 16 = 25$.
  2. $|z| = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$, so $|z|^2 = 25$.
  3. Both sides equal $25$.

Answer: $z\,\overline{z} = |z|^2 = 25$ ✓

Key Points

  • Plot $z = a + ib$ as the point $(a, b)$ in the Argand plane; the conjugate $\overline{z} = a - ib$ is its reflection across the real axis.
  • Modulus $|z| = \sqrt{a^2 + b^2}$ is the distance from the origin, and $z\,\overline{z} = |z|^2$.
  • The multiplicative inverse is $z^{-1} = \dfrac{\overline{z}}{|z|^2}$ for any non-zero $z$.
  • Polar form: $z = r(\cos\theta + i\sin\theta)$ with $r = |z|$ and $\tan\theta = \dfrac{b}{a}$.
  • The principal argument lies in $(-\pi, \pi]$ — always fix it using the quadrant of $(a, b)$.
Tap an option to check your answer0 / 4
Q1.$|3+4i|=$
Explanation: $\sqrt{3^2+4^2}=5$.
Q2.The conjugate of $2-5i$ is:
Explanation: Conjugate flips the sign of the imaginary part.
Q3.$z\bar z$ equals:
Explanation: $z\bar z=|z|^2$.
Q4.$|i|=$
Explanation: $|i|=\sqrt{0^2+1^2}=1$.