1. Theoretical Summary
1.1 Pure Imaginary Unit $i$
The pure imaginary unit, denoted by $i$, is defined as the principal square root of $-1$. It satisfies the fundamental property:
\[ i^2 = -1 \]
Powers of $i$ are cyclic with a period of 4:
\[ i^{4k} = 1, \quad i^{4k+1} = i, \quad i^{4k+2} = -1, \quad i^{4k+3} = -i \quad \text{for } k \in \mathbb{Z} \]
1.2 Complex Number $(x, y)$
A complex number $z$ can be formally defined as an ordered pair of real numbers $(x, y) \in \mathbb{R}^2$. In algebraic form, it is expressed as:
\[ z = x + iy \]
where $x = \text{Re}(z)$ represents the real part and $y = \text{Im}(z)$ represents the imaginary part. The complex conjugate of $z$ is given by $\overline{z} = x - iy$.
1.3 Operations on Complex Numbers
Let $z_1 = (x_1, y_1) = x_1 + iy_1$ and $z_2 = (x_2, y_2) = x_2 + iy_2$.
- Addition: $z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2)$
- Multiplication: $z_1 z_2 = (x_1 x_2 - y_1 y_2) + i(x_1 y_2 + x_2 y_1)$
- Division: $\frac{z_1}{z_2} = \frac{z_1 \overline{z_2}}{|z_2|^2} = \frac{x_1 x_2 + y_1 y_2}{x_2^2 + y_2^2} + i\frac{x_2 y_1 - x_1 y_2}{x_2^2 + y_2^2} \quad (z_2 \neq 0)$
1.4 Trigonometric Form
Any complex number $z = x + iy$ can be expressed in polar or trigonometric form:
\[ z = r(\cos\theta + i\sin\theta) \]
where $r = |z| = \sqrt{x^2 + y^2}$ is the modulus, and $\theta = \arg(z)$ is the argument satisfying $\tan\theta = \frac{y}{x}$. The principal argument is conventionally restricted to $-\pi \lt \theta \leq \pi$.
1.5 Roots of Complex Numbers
By de Moivre's Theorem, the $n$-th roots of a complex number $z = r(\cos\theta + i\sin\theta)$ are given by the $n$ distinct values:
\[ w_k = \sqrt[n]{r} \left( \cos\frac{\theta + 2k\pi}{n} + i\sin\frac{\theta + 2k\pi}{n} \right) \]
where $k = 0, 1, 2, \dots, n-1$.
Example 1
Solve $x^2 + 2x + 5 = 0$Solution
Click to view detailed Solution \[\begin{align*} x^2 + 2x + 5 &= 0\\ x^2+2x+1&=-4\\ (x+1)^2&=-4\\ x+1&=\pm\sqrt 4i\\ x&=-1\pm 2i \end{align*}\] So there are two solutions $x=-1+2i$ and $x=-1-2i.$==============
Example 2
Solve $ x^2 + 2x + 3 = 0 $ and check your answers.Solution
Click to view detailed solution \[ \begin{align*} x^2 + 2x + 3 &= 0 \\ x^2 + 2x &= -3 \\ x^2 + 2x + 1 &= -3 + 1 \\ (x + 1)^2 &= -2 \\ x + 1 &= \pm \sqrt{2}\, i \\ x &= -1 \pm \sqrt{2}\, i \end{align*}\] So two solutions are $ x = -1 + \sqrt{2}\, i $ and $x = -1 - \sqrt{2}\, i .$Exercise 1.1
Question 1
Click to view Example 11. Solve the following equations.
(a) $x^{2}-6x+10=0$
Click to view Solution
\[\begin{align*}
x^{2} - 6x + 10 &= 0 \\
x^{2} - 6x &= -10 \\
x^{2} - 6x + 9 &= -10 + 9 \quad \text{(complete the square: } (-6/2)^2 = 9\text{)} \\
(x - 3)^{2} &= -1 =i^2\\
x - 3 &= \pm i \\
x &= 3 \pm i
\end{align*}\]
$\quad$ Short Answer
$x=3\pm i$
(b) $-2x^{2}+4x-3=0$
Click to view Solution
\[ \begin{align*}
-2x^{2} + 4x - 3 &= 0 \\
x^{2} - 2x + \frac{3}{2} &= 0 \quad \text{(divide by -2)} \\
x^{2} - 2x &= -\frac{3}{2} \\
x^{2} - 2x + 1 &= -\frac{3}{2} + 1 \quad \text{(complete the square: } (-2/2)^2 = 1\text{)} \\
(x - 1)^{2} &= -\frac{3}{2} + \frac{2}{2} \\
(x - 1)^{2} &= -\frac{1}{2} \\
x - 1 &= \pm \sqrt{\frac{1}{2}} i\\
x - 1 &= \pm \frac{i}{\sqrt{2}} \\
x &= 1 \pm \frac{i}{\sqrt{2}}\]
$\quad$ Short Answer
$1 \pm \frac{i}{\sqrt{2}}$
(c) $5x^{2}-2x+1=0$
Click to view Solution
\[\begin{align*}
5x^{2} - 2x + 1 &= 0 \\
x^{2} - \frac{2}{5}x + \frac{1}{5} &= 0 \\
x^{2} - \frac{2}{5}x &= -\frac{1}{5} \\
x^{2} - \frac{2}{5}x + \left(\frac{1}{5}\right)^{2} &= -\frac{1}{5} + \left(\frac{1}{5}\right)^{2} \quad \text{(complete the square: } \left(\frac{-2/5}{2}\right)^{2} = \left(\frac{1}{5}\right)^{2}\text{)} \\
\left(x - \frac{1}{5}\right)^{2} &= -\frac{1}{5} + \frac{1}{25} \\
\left(x - \frac{1}{5}\right)^{2} &= -\frac{5}{25} + \frac{1}{25} \\
\left(x - \frac{1}{5}\right)^{2} &= -\frac{4}{25} \\
x - \frac{1}{5} &= \pm \sqrt{\frac{4}{25}}i \\
x - \frac{1}{5} &= \pm \frac{2i}{5} \\
x &= \frac{1}{5} \pm \frac{2i}{5}
\end{align*}\]
$\quad$ Short Answer
$ \frac{1}{5} \pm \frac{2i}{5}$
(d) $3x^{2}+7x+5=0$
Click to view Solution
\[ \begin{align*}
3x^{2} + 7x + 5 &= 0 \\
x^{2} + \frac{7}{3}x + \frac{5}{3} &= 0 \\
x^{2} + \frac{7}{3}x &= -\frac{5}{3} \\
x^{2} + \frac{7}{3}x + \left(\frac{7}{6}\right)^{2} &= -\frac{5}{3} + \left(\frac{7}{6}\right)^{2} \quad \text{(complete the square: } \left(\frac{7/3}{2}\right)^{2} = \left(\frac{7}{6}\right)^{2}\text{)} \\
\left(x + \frac{7}{6}\right)^{2} &= -\frac{5}{3} + \frac{49}{36} \\
\left(x + \frac{7}{6}\right)^{2} &= -\frac{60}{36} + \frac{49}{36} \\
\left(x + \frac{7}{6}\right)^{2} &= -\frac{11}{36} \\
x + \frac{7}{6} &= \pm \sqrt{\frac{11}{36}} i\\
x + \frac{7}{6} &= \pm \frac{\sqrt{11}\, i}{6} \\
x &= -\frac{7}{6} \pm \frac{\sqrt{11}\, i}{6}
\end{align*}\]
$\quad$ Short Answer
\[-\frac{7}{6} \pm \frac{\sqrt{11}\, i}{6}\]
Question 2
Click to view Example 2Solve the following equations and check your answers.
(a) $x^2-2x+4=0$
Click to view Solution
\[\begin{align*}
x^2 - 2x + 4 &= 0 \\
x^2 - 2x &= -4 \\
x^2 - 2x + 1 &= -4 + 1 \quad \text{(complete the square: } (-2/2)^2 = 1\text{)} \\
(x - 1)^2 &= -3 \\
x - 1 &= \pm \sqrt{3}\, i \\
x &= 1 \pm \sqrt{3}\, i
\end{align*}\]
$\quad$ Short Answer
\[ 1 \pm \sqrt{3}\, i\]
(b) $x^2-4x+5=0$
Click to view Solution
\[ \begin{align*}
x^2 - 4x + 5 &= 0 \\
x^2 - 4x &= -5 \\
x^2 - 4x + 4 &= -5 + 4 \quad \text{(complete the square: } (-4/2)^2 = 4\text{)} \\
(x - 2)^2 &= -1 \\
x - 2 &= \pm i \\
x &= 2 \pm i
\end{align*}\]
$\quad$ Short Answer
$ 2 \pm i$
Question 3
Find the value of $i^n$ for every positive integer $n$, where $i^2=-1, i^3=i^2i, i^4=i^2i^2$, etc.Click to view Solution
Let $n=4q+r,$ for some positive integer $q$ (quotient), $0\le r\le 4,$ (remainder). Then
\[\begin{align*}
i^{n}&=i^{4q+r}\\ &=i^{4q}\times i^r, \text{ where } r=0,1,2,3.\\
&=\left(i^4\right)^q \times i^r\\
&=1^q\times i^r\\
&=i^r
\end{align*}\]
Since $i^0=1,i^1=i,i^2=-1,i^3=-i,$
\[i^n=\left\{ \begin{array}{ll}
1&\text{ when }r=0\\ i&\text{ when }r=1\\ -1&\text{ when }r=2\\ -i&\text{ when }r=3
\end{array}\right.\]