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. Purely real numbers lie on the real axis; purely imaginary numbers lie on the imaginary axis.

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. The conjugate obeys a set of clean properties that are worth knowing as a unit:

PropertyStatement
Conjugate of sum$\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
Conjugate of product$\overline{z_1 z_2} = \overline{z_1}\,\overline{z_2}$
Double conjugate$\overline{(\overline{z})} = z$
Real / imaginary parts$z + \overline{z} = 2\operatorname{Re}(z)$,   $z - \overline{z} = 2i\operatorname{Im}(z)$
Real test$z = \overline{z} \iff z$ is real

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:

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

The modulus has its own list of properties, all of which echo the fact that it measures a length:

PropertyStatement
Non-negative$|z| \ge 0$, with $|z| = 0 \iff z = 0$
Conjugate & negative$|z| = |\overline{z}| = |-z|$
Product$|z_1 z_2| = |z_1|\,|z_2|$
Quotient$\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$   $(z_2 \ne 0)$
Triangle inequality$|z_1 + z_2| \le |z_1| + |z_2|$

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 (or amplitude) 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 the 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}$

Modulus equations as geometry. An equation in $|z|$ describes a locus on the Argand plane. For instance $|z| = r$ is the circle of radius $r$ centred at the origin, $|z - z_0| = r$ is the circle of radius $r$ centred at the point $z_0$, and $|z - z_1| = |z - z_2|$ is the perpendicular bisector of the segment joining $z_1$ and $z_2$ (the points equidistant from both). Reading these geometrically is far faster than expanding into $x$ and $y$.

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: $z\,\overline{z} = |z|^2 = 25$ ✓

7
Worked Example
Find the modulus of $z = \dfrac{1 + i\sqrt{3}}{1 - i\sqrt{3}}$ using the quotient property.
Solution
  1. Use $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$ instead of simplifying first.
  2. $|1 + i\sqrt{3}| = \sqrt{1 + 3} = 2$ and $|1 - i\sqrt{3}| = \sqrt{1 + 3} = 2$.
  3. So the modulus of the quotient is $\dfrac{2}{2} = 1$.

Answer: $|z| = 1$

8
Worked Example
Convert $z = \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right)$ back to the form $a + ib$.
Solution
  1. Here $r = \sqrt{2}$ and $\theta = \dfrac{\pi}{4}$.
  2. $a = r\cos\theta = \sqrt{2}\cdot\dfrac{1}{\sqrt{2}} = 1$.
  3. $b = r\sin\theta = \sqrt{2}\cdot\dfrac{1}{\sqrt{2}} = 1$.

Answer: $z = 1 + i$

9
Worked Example
If $z = 2 + 3i$, find $z + \overline{z}$ and $z - \overline{z}$, and confirm the standard identities.
Solution
  1. $\overline{z} = 2 - 3i$.
  2. $z + \overline{z} = (2 + 3i) + (2 - 3i) = 4 = 2\operatorname{Re}(z) = 2(2)$. ✓
  3. $z - \overline{z} = (2 + 3i) - (2 - 3i) = 6i = 2i\operatorname{Im}(z) = 2i(3)$. ✓

Answer: $z + \overline{z} = 4 = 2\operatorname{Re}(z)$ and $z - \overline{z} = 6i = 2i\operatorname{Im}(z)$

10
Worked Example
Describe the locus of points satisfying $|z - 2| = 3$ on the Argand plane.
Solution
  1. $|z - z_0| = r$ is a circle of radius $r$ centred at $z_0$.
  2. Here $z_0 = 2$ (the point $(2, 0)$) and $r = 3$.
  3. Algebraic check: with $z = x + iy$, $\sqrt{(x-2)^2 + y^2} = 3 \Rightarrow (x-2)^2 + y^2 = 9$.

Answer: A circle of radius $3$ centred at $(2, 0)$, i.e. $(x-2)^2 + y^2 = 9$.

11
Worked Example
Find the modulus and principal argument of $z = -2 - 2i$.
Solution
  1. Modulus: $r = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$.
  2. The point $(-2, -2)$ is in the third quadrant, so $\theta = -\pi + \tan^{-1}\dfrac{2}{2} = -\pi + \dfrac{\pi}{4}$.
  3. $\theta = -\dfrac{3\pi}{4}$.

Answer: $r = 2\sqrt{2},\ \arg(z) = -\dfrac{3\pi}{4}$

12
Worked Example
Verify the product property $|z_1 z_2| = |z_1|\,|z_2|$ for $z_1 = 1 + i$ and $z_2 = 2 - i$.
Solution
  1. Product: $z_1 z_2 = (1 + i)(2 - i) = 2 - i + 2i - i^2 = 3 + i$, so $|z_1 z_2| = \sqrt{9 + 1} = \sqrt{10}$.
  2. Separately: $|z_1| = \sqrt{2}$ and $|z_2| = \sqrt{5}$.
  3. $|z_1|\,|z_2| = \sqrt{2}\cdot\sqrt{5} = \sqrt{10}$.

Answer: Both sides equal $\sqrt{10}$, so $|z_1 z_2| = |z_1|\,|z_2|$. ✓

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.
  • Conjugate properties: $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$, $\overline{z_1 z_2} = \overline{z_1}\,\overline{z_2}$, $z + \overline{z} = 2\operatorname{Re}(z)$, $z - \overline{z} = 2i\operatorname{Im}(z)$.
  • Modulus $|z| = \sqrt{a^2 + b^2}$ is the distance from the origin, and $z\,\overline{z} = |z|^2$.
  • Modulus properties: $|z| = |\overline{z}| = |-z|$, $|z_1 z_2| = |z_1||z_2|$, $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$, and $|z_1 + z_2| \le |z_1| + |z_2|$.
  • 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 $(-\pi, \pi]$ — fix it from the quadrant of $(a, b)$.
  • Modulus equations are loci: $|z| = r$ is a circle at the origin, $|z - z_0| = r$ a circle at $z_0$, and $|z - z_1| = |z - z_2|$ the perpendicular bisector of $z_1 z_2$.
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: $|i|=\sqrt{0^2+1^2}=1$.