Complex Numbers
\[
\mathbb{N} = \{1,2,3,4,\dots\} \text{is the set of } \textbf{natural } \text{numbers used in counting. 1 to } \infty
\
\]
\[
\mathbb{W} = \{0,1,2,3,4,\dots\} \text{is the set of } \textbf{whole } \text{numbers zero to } \infty
\]
\[
\mathbb{Z} = \{\dots,-2,-1,0,1,2,\dots\}
\text{is the set of } \textbf {integers} \text{: positive and negative whole numbers} \]
\[
\begin{aligned}
\mathbb{Q} &\text{ is the set of rational numbers or }\textbf {quotients.}
\\[0.6em]
&\text{These are all numbers which can be expressed as}
\\[0.6em]
&\text{a fraction } \frac{a}{b} \text{ where both } a \text{ and } b \text{ are integers,}
\\[0.6em]
&\text{and } b \text{ is not zero.}
\end{aligned}
\]
\[
\begin{aligned}
\mathbb{R} &\text { is the set of } \textbf {real } \text {numbers } \{ -\infty , \ldots , \infty \}
\\[0.6em]
&\text{This includes all numbers, rational and irrational.}
\end{aligned}
\]
\[
\mathbb{N} \;\subset\; \mathbb{W} \;\subset\; \mathbb{Z} \;\subset\; \mathbb{Q} \;\subset\; \mathbb{R}
\]
\[
\begin{aligned}
\mathbb{C} &\text{ is the set of } \textbf {complex } \text{numbers, } a + b i
\\[0.6em]
&\text{where } a \text{ and } b \text{ are real, and } i \text{ is the imaginary number } \sqrt{-1}.
\end{aligned}
\]
\[
\text{Given } \; z = a + b\,i
\]
\[
a \text{ is called the real part of } z
\]
\[
a = \Re(z)
\qquad\text{or}\qquad
a = \operatorname{Re}(z)
\]
\[
b \text{ is called the imaginary part of } z
\]
\[
b = \Im(z)
\qquad\text{or}\qquad
b = \operatorname{Im}(z)
\]
This is reflection in the real axis.
\[
z = a + b i
\quad\text{has complex conjugate}\quad
\overline{z} = a - b i
\]
\[
z\,\overline{z}
= (a + b i)(a - b i)
= a^{2} + b^{2}
\]
Example
\[
\text{Given } z = 2 - 3i
\]
\[
\text{Find }
\quad
\text{a) } \overline{z}
\qquad\qquad\qquad
\text{b) } z\,\overline{z}
\]
\[
\text{a) } \overline{z} = 2 + 3i
\]
\[
\text{b) } z\,\overline{z}
= (2 - 3i)(2 + 3i)
= 4 + 9
= 13
\]
\[
\text{Given }
\quad z_{1} = a + b i
\quad\text{and}\quad
z_{2} = c + d i
\]
\[
z_{1} + z_{2}
= (a + c) + (b + d)i
\]
\[
z_{1} - z_{2}
= (a - c) + (b - d)i
\]
\[
z_{1}z_{2}
= (a + b i)(c + d i)
\]
\[
= (ac - bd) + (bc + ad)i
\]
\[
\frac{z_{1}}{z_{2}}
= \frac{a + b i}{c + d i}
\]
\[
= \frac{(a + b i)(c - d i)}{c^{2} + d^{2}}
\]
\[
= \frac{(ac + bd) + (bc - ad)i}{c^{2} + d^{2}}
\]
Example
Solve the equation \(x^2 - 2x + 5 = 0\)
\[
x^{2} - 2x + 5 = 0
\]
\[
x = \frac{-b \pm \sqrt{\,b^{2} - 4ac\,}}{2a}
\]
\[
= \frac{2 \pm \sqrt{(-2)^{2} - 20}}{2}
\]
\[
= \frac{2 \pm \sqrt{-16}}{2}
\]
\[
= \frac{2 \pm 4i}{2}
\]
\[
= 1 \pm 2i
\]
Example
\[
\text{Given }
\quad z_{1} = 2 - 3i
\quad\text{and}\quad
z_{2} = 1 + i
\]
\[
\text{Find}
\]
\[
\text{a) } z_{1} + z_{2}
\]
\[
\text{b) } z_{1} - z_{2}
\]
\[
\text{c) } z_{1} z_{2}
\]
\[
\text{d) } \frac{z_{1}}{z_{2}}
\]
\[
\text{a) } z_{1} + z_{2}
= (2 + 1) + (-3 + 1)i
\]
\[
= 3 - 2i
\]
\[
\text{b) } z_{1} - z_{2}
= (2 - 1) + (-3 - 1)i
\]
\[
= 1 - 4i
\]
\[
\text{c) } z_{1}z_{2}
= (2 - 3i)(1 + i)
\]
\[
= 2 - 3i + 2i + 3
\]
\[
= 5 - i
\]
\[ \text{d) }
\frac{z_{1}}{z_{2}}
= \frac{\,2 - 3i\,}{\,1 + i\,}
\]
\[
= \frac{2 - 3i}{1 + i}
\times
\frac{1 - i}{1 - i}
\]
\[
= \frac{(2 - 3i)(1 - i)}{(1 + i)(1 - i)}
\]
\[
= \frac{2 - 3i - 2i - 3}{1^{2} + (-1)^{2}}
\]
\[
= \frac{-1 - 5i}{2}
\]
Alternatively:
\[
\text{d)}
\]
\[
z_{1} = 2 - 3i
\quad\text{and}\quad
z_{2} = 1 + i
\]
\[
a = 2 \qquad c = 1
\]
\[
b = -3 \qquad d = 1
\]
\[
\frac{z_{1}}{z_{2}}
=
\frac{(ac + bd) + (bc - ad)i}{c^{2} + d^{2}}
\]
\[
=
\frac{(2\cdot1 + (-3)\cdot1) + \big((-3)\cdot1 - 2\cdot1\big)i}{1^{2} + (-1)^{2}}
\]
\[
=
\frac{(2 - 3) + (-3 - 2)i}{1 + 1}
\]
\[
=
\frac{-1 - 5i}{2}
\]
Example
\[
\text{Given } z = 2 - 3i
\quad\text{Find } z^{-1}
\]
\[
z = 2 - 3i
\]
\[
\Rightarrow\quad
z^{-1} = \frac{1}{\,2 - 3i\,}
\]
\[
= \frac{1}{2 - 3i}
\times
\frac{2 + 3i}{2 + 3i}
\]
\[
= \frac{2 + 3i}{4 + 9}
\]
\[
= \frac{2 + 3i}{13}
\]
\(z = x + iy\) represented as point \(P(x,y)\).
Points on x‑axis are real; points on y‑axis are imaginary.
Representation as vector
The modulus \(r\) is the length of the vector.
\[
r
= \overrightarrow{OP}
= |z|
\]
\[
r = \sqrt{x^{2} + y^{2}}
\]
\[
\therefore\ |z| = \sqrt{x^{2} + y^{2}}
\]
The argument is the angle of rotation.
\[
\sin\theta = \frac{y}{r}
\qquad\qquad\qquad
\cos\theta = \frac{x}{r}
\]
\[
\Rightarrow\;
y = r\sin\theta
\qquad\qquad
x = r\cos\theta
\]
\[
\tan\theta
= \frac{\sin\theta}{\cos\theta}
= \frac{y}{x}
\]
\[
\Rightarrow\;
\theta = \tan^{-1}\!\left(\frac{y}{x}\right)
\]
\[
\text{Arg }z
= \tan^{-1}\!\left(\frac{y}{x}\right) + 2n\pi
\quad\text{radians}
\]
Principal argument lies in \(-\pi \lt \theta \le \pi\).
Example
Find modulus and argument of \(z = 3 + 4i\).
\[
|z| = \sqrt{x^{2} + y^{2}}
\]
\[
= \sqrt{3^{2} + 4^{2}}
\]
\[
= \sqrt{25}
\]
\[
= 5
\]
\[
\text{Arg }z
= \tan^{-1}\!\left(\frac{y}{x}\right)
\]
\[
= \tan^{-1}\!\left(\frac{4}{3}\right)
\]
\[
= 0.927 + n\pi \;\text{rads}
\]
\[
\text{Since the point is in the first quadrant, } n = 0
\]
\[
\arg z = 0.927 \;\text{rads}
\]
Given that \[ z = a + b i \] , find the locus of each expression:
a) \( |z| = 3 \)
b) \(|z -5| = 8\)
c) \(|z -5i| = 8\)
d) \(|2z -3i| = 5\)
\[
\text{a) } |z| = 3
\]
\[
\Rightarrow\quad \sqrt{x^{2} + y^{2}} = 3
\]
\[
\Rightarrow\quad x^{2} + y^{2} = 9
\]
\[
\text{This describes a circle with centre } (0,0) \text{ and radius } 3
\]
\[
\text{b) } |\,z - 5\,| = 8
\]
\[
\Rightarrow\quad |\,x - 5 + iy\,| = 8
\]
\[
\Rightarrow\quad \sqrt{(x - 5)^{2} + y^{2}} = 8
\]
\[
\Rightarrow\quad (x - 5)^{2} + y^{2} = 64
\]
\[
\text{This describes a circle with centre } (5,0)
\]
\[
\text{radius } 8
\]
\[ \text{c) } |\,z - 5i\,| = 8
\]
\[
\Rightarrow\quad |\,x + iy - 5i\,| = 8
\]
\[
\Rightarrow\quad \sqrt{x^{2} + (y - 5)^{2}} = 8
\]
\[
\Rightarrow\quad x^{2} + (y - 5)^{2} = 64
\]
\[
\text{This describes a circle with centre } (0,5)
\]
\[
\text{radius } 8
\]
\[ \text{d) } |\,2z - 3i\,| = 5
\]
\[
\Rightarrow\quad |\,2(x + iy) - 3i\,| = 5
\]
\[
\Rightarrow\quad |\,2x + 2iy - 3i\,| = 5
\]
\[
\Rightarrow\quad |\,2x + i(2y - 3)\,| = 5
\]
\[
\Rightarrow\quad \sqrt{(2x)^{2} + (2y - 3)^{2}} = 5
\]
\[
\Rightarrow\quad 2x^{2} + (2y - 3)^{2} = 25
\]
This describes an ellipse with centre \((0, 3/2)\).
\[
y = r\sin\theta
\qquad\qquad
x = r\cos\theta
\]
\[
z = x + iy
\quad\text{becomes}
\]
\[
z = r\cos\theta + r\sin\theta\, i
\]
\[
= r(\cos\theta + i\sin\theta)
\]
\[
z = r(\cos\theta + i\sin\theta)
\]
Example
Express \(z = 3 + 4i\) in polar form.
\[
\text{Modulus:}
\]
\[
r = |z| = \sqrt{x^{2} + y^{2}}
\]
\[
= \sqrt{3^{2} + 4^{2}}
\]
\[
= \sqrt{25}
\]
\[
= 5
\]
\[
\text{Argument:}
\]
\[
\theta = \arg z = \tan^{-1}\!\left(\frac{y}{x}\right)
\]
\[
= \tan^{-1}\!\left(\frac{4}{3}\right)
\]
\[
= 0.927
\]
\[
z = r(\cos\theta + i\sin\theta)
\]
\[
= 5(\cos 0.927 + i\sin 0.927)
\]
\[
|z|^{2} = z\,\overline{z}
\]
\[
|\overline{z}| = |z|
\]
\[
\left|\frac{1}{z}\right| = \frac{1}{|z|}
\]
\[
|\,z_{1}z_{2}\,| = |z_{1}|\,|z_{2}|
\]
\[
\left|\frac{z_{1}}{z_{2}}\right|
= \frac{|z_{1}|}{|z_{2}|}
\]
\[
z^{-1} = \frac{\overline{z}}{|z|^{2}},\qquad z\neq 0
\]
\[
\Arg\!\left(\frac{1}{z}\right)
= -\,\Arg(z)
\]
\[
\Arg(z_{1}z_{2})
= \Arg(z_{1}) + \Arg(z_{2})
\]
\[
\Arg\!\left(\frac{z_{1}}{z_{2}}\right)
= \Arg(z_{1}) - \Arg(z_{2})
\]
\[
z^{n}
= r^{\,n}\!\left(\cos(n\theta) + i\sin(n\theta)\right)
\]
Example
Compute \(z^5\) for \(z = 3 + 4i\).
\[
|z| = \sqrt{x^{2} + y^{2}}
\]
\[
= \sqrt{3^{2} + 4^{2}}
\]
\[
= \sqrt{25}
\]
\[
= 5
\]
\[
\begin{aligned}
\theta &= \text{arg}\,z = \tan^{-1}\!\left(\frac{y}{x}\right) \\
&= \tan^{-1}\!\left(\frac{4}{3}\right) \\
&= 0.927
\end{aligned}
\]
\[
\begin{aligned}
z &= r\bigl(\cos\theta + i\sin\theta\bigr) \\
&= 5\bigl(\cos(0.927) + i\sin(0.927)\bigr)
\end{aligned}
\]
\[
\begin{aligned}
z^{5}
&= \left[\,5\bigl(\cos(0.927) + i\sin(0.927)\bigr)\right]^{5} \\[6pt]
&= 5^{5}\,\bigl(\cos(5\times 0.927) + i\sin(5\times 0.927)\bigr) \\[6pt]
&= 3125\,\bigl(\cos(4.635) + i\sin(4.635)\bigr) \\[6pt]
&= 3125\,\bigl(-0.0773 - 0.997\,i\bigr) \\[6pt]
&= -241.5625 - 3115.625\,i \\[6pt]
&= -242 - 3116\,i \quad\text{(nearest integer)}
\end{aligned}
\]
Roots of a complex number
\[
\begin{aligned}
z &= r\bigl(\cos\theta + i\sin\theta\bigr) \\[6pt]
\text{has } n \text{ solutions } \quad z_k^{\,n} &= z \\[10pt]
z_k &= r^{\,1/n}\!
\left(
\cos\!\left(\frac{\theta + 2k\pi}{n}\right)
+ i\sin\!\left(\frac{\theta + 2k\pi}{n}\right)
\right) \\[6pt]
k &= 0,1,2,3,\ldots,n-1
\end{aligned}
\]
Example
Solve the equation
\[
z^{4} = -3 + 3\sqrt{3}\,i
\]
\[
\begin{aligned}
z^{4} &= -3 + 3\sqrt{3}\,i \\[8pt]
\left|z^{4}\right|
&= \sqrt{(-3)^{2} + \left(3\sqrt{3}\right)^{2}} \\[8pt]
&= \sqrt{36} \\[8pt]
&= 6
\end{aligned}
\]
\[
\begin{aligned}
\theta &= \arg z = \tan^{-1}\!\left(\frac{y}{x}\right) \\[6pt]
&= \tan^{-1}\!\left(\frac{3\sqrt{3}}{-3}\right) \\[6pt]
&= \tan^{-1}(-\sqrt{3}) \\[6pt]
&= \frac{2\pi}{3}
\end{aligned}
\]
\[
\begin{aligned}
z^{4} &= -3 + 3\sqrt{3}\,i \\[10pt]
\text{has solutions of the form} \\[6pt]
z &= 6^{1/4}\!\left(
\cos\!\left(
\frac{1}{4}\left(\frac{2\pi}{3} + 2k\pi\right)
\right)
+ i\sin\!\left(
\frac{1}{4}\left(\frac{2\pi}{3} + 2k\pi\right)
\right)
\right),
\quad k = 0,1,2,3
\end{aligned}
\]
\[
\begin{aligned}
\text{when } k &= 0 \\[6pt]
z &= 6^{1/4}\!\left(
\cos\!\left(\frac{\pi}{6}\right)
+ i\sin\!\left(\frac{\pi}{6}\right)
\right)
\end{aligned}
\]
\[
\begin{aligned}
\text{when } k &= 1 \\[6pt]
z &= 6^{1/4}\!\left(
\cos\!\left(\frac{2\pi}{3}\right)
+ i\sin\!\left(\frac{2\pi}{3}\right)
\right)
\end{aligned}
\]
\[
\begin{aligned}
\text{when } k &= 2 \\[6pt]
z &= 6^{1/4}\!\left(
\cos\!\left(\frac{7\pi}{6}\right)
+ i\sin\!\left(\frac{7\pi}{6}\right)
\right)
\end{aligned}
\]
\[
\begin{aligned}
\text{when } k &= 3 \\[6pt]
z &= 6^{1/4}\!\left(
\cos\!\left(\frac{19\pi}{6}\right)
+ i\sin\!\left(\frac{19\pi}{6}\right)
\right)
\end{aligned}
\]
If a polynomial has an unreal root, it has complex conjugate roots.
\[
r(\cos\theta + i\sin\theta)
\qquad\text{and}\qquad
r(\cos\theta - i\sin\theta)
\]
A polynomial of degree \(n\) has \(n\) complex roots.
Example
Find the roots of \(z^3 - 6z^2 + 13z - 20 = 0\), given \(z = 1 + 2i\) is a root.
If \(1 + 2i\) is a root, so is \(1 - 2i\).
\[
(z - 1 - 2i)(z - 1 + 2i)
= z^{2} - 2z + 5
\]
