1.1 Complex Arithmetic

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We perform arithmetic operations on complex numbers.

1 Complex Numbers

What is a complex number? The core component of a complex number is the imaginary number $i = \sqrt {-1}$ .

A complex number is a number of the form $a+bi$ where $a,b \in \mathbb {R}$ . The set of all complex numbers is $\mathbb {C}= \{a+bi \,|\, a,b \in \mathbb {R}\}$ . The real part of $\,a+bi\,$ is ‘ $a$ ’ and the imaginary part of $\,a+bi\,$ is ‘ $b$ ’.

The imaginary part of a complex number is a real number.

We use the symbols $\Rep$ and $\Imp$ to denote the real and imaginary parts of a complex number: $\Rep (a+bi) = a \qquad \mbox {and} \qquad \Imp (a+bi) = b$

Example 1 The following are complex numbers: $2+3i, \; -2i$ and $\sqrt 2$ . Their real and imaginary parts are: \begin{align*} (i)& \quad \Rep (2+3i) = 2 \quad \mbox{and} \quad \;\Imp (2+3i) = 3 \\ (ii)& \quad \Rep (-2i) = 0 \quad \;\;\; \mbox{and} \quad \;\Imp (-2i) = -2 \\ (iii)& \quad \Rep \left (\sqrt 2\right ) = \sqrt 2 \quad \;\mbox{and} \quad \;\Imp \left (\sqrt 2\right ) = 0 \end{align*}

A real number is a complex number, i.e., $\mathbb {R} \subset \mathbb {C}$ . Moreover, the imaginary part of a real number is $0$ .

If the real part of a complex number is $0$ , then we say that the complex number is purely imaginary.

(Problem 1a) Find the real and imaginary parts of the following complex numbers: $-3i,\; \pi$ and $-2+i$ . \begin{align*} (i) \quad &\Rep (-3i) = \answer{0} \qquad \quad \; \mbox{and} \quad \; \Imp (-3i) = \answer{-3}\\ (ii) \quad &\Rep (\pi ) = \answer{\pi } \qquad \qquad \; \mbox{and} \quad \;\Imp (\pi ) = \answer{0}\\ (iii) \quad &\Rep (-2+i) = \answer{-2} \qquad \mbox{and} \quad \; \Imp (-2+i) = \answer{1}\\ \end{align*}

(Problem 1b) Label each complex number as either real, purely imaginary or neither:
(i) $-3i$ realpurely imaginary neither (ii) $\pi$ realpurely imaginary neither (iii) $-2+i$ realpurely imaginary neither (iv) $0$ realpurely imaginary neitherboth!

Two complex numbers are equal if their real and imaginary parts are both equal.

Example 2 The complex numbers $\frac 12 - \frac 13 i$ and $0.5 - 0.{\overline 3}i$ are equal since $\frac 12 = 0.5$ and $-\frac 13 = -0.{\overline 3}$ .

(Problem 2) Which of the following complex numbers are equal?

$1 + 0.125i$ and $0.{\overline 9} + \frac 18 i$ $\cos \left (\tfrac {\pi }{6}\right ) + i\sin \left (\tfrac {\pi }{6}\right )$ and $\frac {\sqrt 3}{2} + \frac 12 i$ $e + \pi i$ and $2.72 + 3.14i$

2 Addition of Complex Numbers

Complex numbers are added by adding the real and imaginary parts independently: $(a+bi) + (c+di) = (a+c) + (b+d)i$

Complex addition is both commutative and associative. This is a consequence of the fact that the addition of real numbers is both commutative and associative.

Example 3 Compute the following sums: \begin{align*} (i)& \quad (3+4i) + (5+6i) = 8+10i \\ (ii)& \quad (3+2i) + (3-2i) = 6 \\ (iii)& \quad (4+i) - (4-i) = 2i \end{align*}

(Problem 3) Compute the following sums: \begin{align*} (i)& \quad (2+3i) + (4+5i) = \answer{6+8i} \\ (ii)& \quad (5+2i) + (5-2i) = \answer{10}\\ (iii)& \quad (6-3i) - (6+3i) = \answer{-6i} \end{align*}

3 Multiplication of Complex Numbers

Complex numbers are multiplied in the same way as binomials, using the fact that $i^2 = -1$ : $(a+bi)(c+di) = ac + adi + bci + (bi)(di) = (ac-bd) + (ad+bc)i$

Complex multiplication is both commutative and associative since the multiplication of real numbers is also both commutative and associative.

Example 4 Compute the following products: \begin{align*} (i)& \quad (3+4i) (5+6i) = 15 + 18i + 20i - 24 = -9 + 38i \\ (ii)& \quad (3+2i) (3-2i) = 9-6i+6i+4=13 \\ (iii)& \quad 5(2i) = 10i \end{align*}

(Problem 4) Compute the following products: \begin{align*} (i)& \quad (2+3i) (4+5i) = \answer{-7+22i}\\ (ii)& \quad (3-i) (2-3i) = \answer{3-11i}\\ (iii)& \quad (4+i) (4-i) = \answer{17}\\ (iv)& \quad -4(3i)= \answer{-12i} \end{align*}

The Powers of $i$

An alternative to stating that $i = \sqrt {-1}$ is to say $i^2 = -1$ . This leads us to compute the powers of $i$ .

Example 5 Compute the each of the following powers: $i^3, i^4$ and $i^{75}$ . \begin{align*} (i)& \quad i^3 = (i^2)(i) = (-1)(i) =-i\\ (ii)& \quad i^4 = (i^2)(i^2) = (-1)(-1) = 1\\ (iii)& \quad i^{75} = (i^{72})(i^3) = (i^4)^{18} (i^3) = i^3 = -i \end{align*}

Note that in part $(iii)$ , a large power of $i$ was written in terms of $i^4$ which is $1$ . In general $i^{4n+k} = i^k$ , where $k$ and $n$ are integers.

(Problem 5) Compute the following powers of $i$ : \begin{align*} (i)& \quad i^5 = \answer{i}\\ (ii)& \quad i^{50} = \answer{-1}\\ (iii)& \quad i^{103}= \answer{-i} \end{align*}

5 The Complex Conjugate

Complex Conjugate The complex conjugate of a complex number $a+bi$ is defined as $\overline {a+bi} = a-bi$

Example 6 Find the complex conjugate: \begin{align*} (i)& \quad \overline{2+3i} = 2-3i \\ (ii)& \quad \overline{1-i} = 1+i \\ (iii)& \quad \overline{75} = 75 \\ (iv)& \quad \overline{4i} = -4i \end{align*}

(Problem 6) Find the complex conjugate: \begin{align*} (i)& \quad{\overline{3+2i}} = \answer{3-2i} \\ (ii)& \quad{\overline{4-5i}} = \answer{4+5i}\\ (iii)& \quad{\overline{26}}= \answer{26}\\ (iv)& \quad{\overline{-7i}}= \answer{7i} \end{align*}

If we multiply a complex number by its conjugate, we get a real number: $(a+bi) \overline {(a+bi)} = (a+bi)(a-bi) = a^2 - (bi)^2 = a^2 +b^2$

Example 7 Multiply the complex conjugates: \begin{align*} (i)& \quad (2+5i) \overline{(2+5i)} = (2+5i)(2-5i) = 2^2 + 5^2 = 29\\ (ii)& \quad (1-i) \overline{(1-i)} = (1-i)(1+i) = 1^2 + (-1)^2 = 2 \end{align*}

(Problem 7) Multiply the complex conjugates: \begin{align*} (i)& \quad (6+2i) \overline{(6+2i)} = \answer{40} \\ (ii)& \quad (3-4i) \overline{(3-4i)} = \answer{25} \end{align*}

6 Division of Complex Numbers

To divide complex numbers, we use the complex conjugate as follows: $\frac {a+bi}{c+di} = \frac {a+bi}{c+di}\cdot \frac {c-di}{c-di} = \frac {(a+bi)(c-di)}{c^2 + d^2} = \left (\frac {ac+bd}{c^2 + d^2}\right ) + \left (\frac {bc-ad}{c^2 + d^2}\right )i$

Example 8 Divide the complex numbers: $\frac {3+2i}{4-5i}$ Begin by multiplying the numerator and denominator by the conjugate of the denominator, which is $\overline {4-5i} = 4+5i$ . We have \begin{align*} \frac{3+2i}{4-5i} &= \frac{3+2i}{4-5i} \cdot \frac{4+5i}{4+5i} \\[9 pt] &= \frac{(3+2i)(4+5i)}{(4-5i)(4+5i)}\\[9pt] &= \frac{(3+2i)(4+5i)}{4^2 +(-5)^2}\\[8pt] &= \frac{2+23i}{41}\\[7pt] &= \frac{2}{41} + \frac{23}{41}i \end{align*}

(Problem 8) Divide the complex numbers: \begin{align*} (i)& \quad \frac{7+3i}{8+5i} = \answer{\frac{71}{89} -\frac{11}{89}i}\\ (ii)& \quad \frac{1+i}{1-i} = \answer{i}\\ (iii)& \quad \frac{4+i}{i} = \answer{1-4i}\\ (iv)& \quad \frac{1}{6+8i} = \answer{\frac{6}{100} - \frac{8}{100}i} (v)&\quad \frac{1}{i} = \answer{-i} \end{align*}

(Problem 9) Find the following quantities: \begin{align*} (i)& \quad \Rep \left [(4+3i)\overline{(2+i)}\right ] = \answer{11}\\ (ii)& \quad \Imp \left [(1+i)^3\right ] = \answer{2}\\ (iii)& \quad \overline{\left (\frac{1}{i^7}\right )} = \answer{-i}\\ (iv)& \quad \overline{\overline{(5-2i)}} = \answer{5-2i} \end{align*}

Here is a video solution to a problem on dividing complex numbers:

2024-09-27 14:05:43