Title: Default for 10.1 Complex Arithmetic
Course: HELM Workbooks 2023-2024

What is a Complex Number?

We assume that you are familiar with the properties of the real numbers; examples are:

\[1,\ -2,\ {3\over {10}},\ 2.634,\ -3.111,\ \pi ,\ e ,\ \sqrt{2}.\]

We all know how to add, subtract and multiply such numbers. We can also divide by any number which is non-zero. We are aware that the numbers can be positive or negative or zero and also aware of their geometrical interpretation as being represented by points on a ‘real’ axis known as a number line as in the figure below.

[Image]

The real axis is a line with a direction (usually chosen to be from left to right) indicated by an arrow. We shall refer to this as the \(x\)-axis. On this axis, we select a point, arbitrarily, and refer to this as the O. The origin, where zero is located, distinguishes positive numbers from negative numbers: the numbers to the right of the origin are the positive numbers and the numbers to the left of the origin are the negative numbers. Thus we can ‘locate’ the numbers in our example.

[Image]

We can formalise the algebra of real numbers into a set of rules which they obey.

So if \(x_1,x_2\) and \(x_3\) are any three real numbers then we know that, in particular:

\[x_1+x_2=x_2+x_1\qquad x_1+(x_2+x_3)=(x_1+x_2)+x_3\]

\[1\times x_1=x_1\qquad 0\times x_1=0\]

\[x_1\times x_2=x_2\times x_1\qquad x_1\times (x_2+x_3)=x_1\times x_2+x_1\times x_3\]

Also, in multiplication we are familiar with the elementary rules: \(\qquad\text{(positive)}\times\text{(positive)}=\text{positive}\qquad\qquad\ \; \text{(positive)}\times\text{(negative)}=\text{negative}\) \(\qquad\text{(negative)}\times\text{(positive)}=\text{negative}\qquad\qquad \text{(negative)}\times\text{(negative)}=\text{positive}\)

It follows that if \(x\) represents any real number then \[\qquad x^2\ge 0\] in words, the square of a real number is always non-negative.

Here we will consider a kind of number (a generalisation of a real number) whose square is not necessarily positive (and not necessarily real either). However, it is important to get to know how to handle complex numbers before using them in calculations. This will not be difficult as the new set of rules is, in fact, precisely the same set of rules obeyed by the ‘real’ numbers. These new numbers are called complex numbers.

A complex number is an ordered pair of real numbers, usually denoted by \(z\) or \(w\) etc. So if \(a,b\) are real numbers then we designate a complex number through: \[\qquad z=a+i b\] where \(i\) is a symbol obeying the rule\[\qquad i^2=-1.\] For simplicity we shall assume we can write \[\qquad i=\sqrt{-1}.\]Often, particularly in electrical engineering applications, the symbol \(\text{j}\) is used instead of \(i.\) Also note that, conventially, examples of actual complex numbers such as \(2+3i\) are written like this and not \(2+i 3\).

Consider for a moment \(\sqrt{-4}\). Clearly this can't be a real number since we would like \((\sqrt{-4})^2=-4\), and the square of any real number is non-negative. Using the rule that \(\sqrt{-1}=i\) we can follow familiar algebraic rules and write \[\sqrt{-4}=\sqrt{(4)(-1)}=\sqrt{4}\sqrt{-1}=2i,\] a complex number.

If \(z=a+i b\) is any complex number then the complex conjugate of \(z\) is denoted by \(z^*\) and is defined by \[z^*=a-i b.\](Sometimes the notation \(\bar z\) is used instead of \(z^*\) to denote the conjugate).

For example if \(z=2-3i\) then \(z^*=2+3i\).

If \(z\) is entirely real then \(z^*=z\) whereas if \(z\) is entirely imaginary then \(z^*=-z\). E.g. if \(z=17i\) then \(z^*=-17i\).

In fact the following relationships are easily obtained: \[\qquad \text{Re}(z)={z+z^*\over 2}\qquad \text{and}\qquad \text{Im}(z)={i(z^*-z)\over {2}}\]

Complex numbers are added, subtracted, multiplied and divided in much the same way as these operations are carried out for real numbers.

Exercise 1

If \(z=[value]\) find expressions for

1. \(\textrm{Re}(z^*) =\)
2. \(\textrm{Im}(i(z^*-z)) =\)
Feedback: 1. We are given \(z=[value]\). Therefore the complex conjugate is \(z^* = [value]\). So \(\textrm{Re}(z^*)\) = [value]. 2. We can then calculate \(z^* - z = [value]\). Therefore \( i (z^* - z) = [value] \). Therefore \( \textrm{Im} (\text{i}(z^*-z)) = [value] \).

Addition and subtraction of complex numbers

Let \(z\) and \(w\) be any two complex numbers where \(z=a+i b\) and \(w=c+i d\) then \[z+w=(a+c)+i(b+d),\] \[z-w=(a-c)+i(b-d).\] For example if \(z=2-3i\), \(w=-4+2i\) then \[z+w=\{2+(-4)\}+\{(-3)+2\}i =-2-i,\] \[z-w=\{2-(-4)\}+\{(-3)-2\}i =6-5i\]

Multiplying complex numbers

In multiplication we proceed using an obvious approach: again consider any two complex numbers \(z=a+i b\) and \(w=c+i d\). Then \[\begin{aligned} zw&=(a+i b)(c+i d)\\ &=ac+aid+ibc+i^2bd\qquad\end{aligned}\] obtained in the usual way by multiplying all the terms in one bracket by all the terms in the other bracket. Now we use the fundamental relation \(i^2=-1\) so that \[\begin{aligned} zw&=ac+ai d+i bc-bd\\ &=ac-bd+i(ad+bc)\qquad\end{aligned}\] where we have re-grouped terms with the ‘\(i\)’ symbol and terms without the ‘\(i\)’ symbol separately. These are the real and imaginary parts of the product \(zw\) respectively. A numerical example will confirm the approach. If \(z=2-3i\) and \(w=-4+2i\) then \[\begin{aligned} zw&=(2-3i)(-4+2i)\\ &=2(-4)+2(2i)-3i(-4)-3i(2i)\qquad\\ &=-8+4i+12i-6i^2\\ &=-8+16i+6\\ &=-2+16i\end{aligned}\]

Exercise
If \(z=[value]\) and \(w=[value]\) find expressions for

1. \([value]=\)
2. \(|[value]| = \)
3. \([value]= \)
Feedback: Given \(z=[value]\) and \(w=[value]\). Then 1. \([value]=[value]([value])+[value]([value])=[value].\) 2. \(|[value]|=|[value]([value])-[value]([value])|=[value].\) 3. \([value]=[value][value][value]=[value].\)

In general the square of a complex number is not necessarily a positive real number; it may not even be real at all. For example if \(z=-2+i\) then \[\qquad z^2=(-2+i)^2=4-4i+i^2=4-4i-1=3-4i.\] However, the product of a complex number with its conjugate is always a non-negative real number. If \(z=a+i b\) then \[\begin{aligned} zz^*&=(a+i b)(a-i b)\\ &=a^2-a(i b)+(i b)a-i^2b^2\\ &=a^2-i^2b^2\\ &=a^2+b^2\qquad \text{since}\ \ i^2=-1 \qquad\end{aligned}\] For example, if \(z=2+i\) then \[\qquad zz^*=(2+i)(2-i)=4+1=5.\]

Dividing complex numbers

Here we consider the operation of dividing one complex number \(z=a+i b\) by another, \(w=c+i d\): \[\qquad {z\over w}={a+i b\over {c+i d}}\] We wish to simplify the right-hand side into the standard form of a complex number : \[\text{(Real part)}\ +i\ \text{(Imaginary part)}\] or the equivalent: \[\text{(Real part)}\ +\ \text{(Imaginary part)}\ i\] To do this we multiply ‘top and bottom’ by the complex conjugate of the bottom (the denominator), that is, by \(c-i d\) (this is called rationalising): \[\qquad {z\over w}={a+i b\over {c+i d}}={a+i b\over {c+i d}}\times {c-i d\over {c-i d}}\] and then carry out the multiplication, top and bottom: \[\begin{aligned} {z\over w}&={(ac+bd)+i(bc-ad)\over {c^2+d^2}}\\ &=\left({ac+bd\over {c^2+d^2}}\right)+i\left({bc-ad\over {c^2+d^2}}\right)\qquad\end{aligned}\] which is now in the required form. The reason for rationalising is to get a real number in the denominator since a complex number divided by a real number is easy to evaluate.

Example

Find \(\displaystyle{{{z\over w}}}\) if \(z=2-3i\) and \(w=2+i\).

Solution

\[\begin{aligned} {z\over w}={2-3i\over {2+i}} &={(2-3i)\times (2-i)\over {(2+i)\times (2-i)}}\qquad \text{rationalising}\\&={4-3+i(-6-2)\over {4+1}}\qquad \text{multiplying out}\qquad\qquad\qquad\qquad\qquad\qquad\\ &={1\over 5}-{8\over 5}i\qquad \text{dividing through}\end{aligned}\]

Exercise

If \(z=[value]\), \(w=[value]\) find expressions (in standard Cartesian form) for
1. \(z-3w=\)
2. \(zw^*=\)
3. \( \displaystyle{\left(\frac{z}{w}\right)^* = }\)
4. \(\displaystyle{\left\vert\frac{z}{w}\right\vert= }\)
Feedback: If \(z=[value]\), \(w=[value]\), 1. \(z-3w=([value])-3*([value])=[value]\). 2. \(zw^*=([value])([value])^*=[value]\). 3. \(\left(\dfrac{z}{w}\right)^*=\left(\dfrac[value][value]\right)^*=\displaystyle{[value]}\). 4. \( \left\vert \dfrac{z}{w}\right\vert= \left\vert \dfrac[value][value] \right\vert=\displaystyle{[value]}\).

Exercise

Find \(z\) such that \(zz^*+3(z-z^*)=[value]\)
Feedback: Let \( z=a+ib \) and so, the conjugate of \(z\) is \(z^*=a-ib\). Now we have \[ zz^*+3(z-z^*)=(a+ib)(a-ib)+3((a+ib)-(a-ib))=[value]\]. Simplifying this and equating the real and complex parts, we get \[ a^2+b^2+6ib=[value]\Rightarrow a=[value] or [value], b=[value].\]

Solutions to polynomial equations

With the introduction of complex numbers we can now obtain solutions to those polynomial equations which may have real solutions, complex solutions or a combination of real and complex solutions. For example, the simple quadratic equation \(x^2+16=0\) can be rearranged as \(x^2=-16\) and then taking square roots we see that \(x=\pm\sqrt{-16}=\pm4\sqrt{-1}=\pm 4i\) where we are replacing \(\sqrt{-1}\) by the symbol ‘\(i\)’.

This approach can be extended to the general quadratic equation \(ax^2+bx+c=0\) with roots \(\displaystyle{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}\) so that if for example \(3x^2+2x+2=0\) then solving for \(x\):

\[\begin{aligned} x&={-2\pm \sqrt{4-4(3)(2)}\over {2(3)}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ &={-2\pm \sqrt{-20}\over {6}}= {-2\pm i\sqrt{20}\over {6}} \\\end{aligned}\]

so, \(\large(\)as \(\displaystyle{\frac{\sqrt{20}}{6}=\frac{2\sqrt{5}}{6}=\frac{\sqrt{5}}{3}}\)\(\large)\), the two roots are \(\displaystyle{-{\frac13} +\frac{\sqrt{5}}{3}i}\) and \(\displaystyle{-{\frac13} -\frac{\sqrt{5}}{3}i}\).

In this example we see that the two solutions (roots) are complex conjugates of each other. In fact this will always be the case if the polynomial equation has coefficients: that is, if any complex roots occur they will always occur in complex conjugate pairs.

Example

Given that \(x=3-2i\) is one root of the cubic equation \(x^3-7x^2+19x-13=0\), find the other two roots.

Solution

Since the coefficients of the equation are real and \(3-2i\) is a root then its complex conjugate \(3+2i\) is also a root which implies that \(x-(3-2i)\) and \(x-(3+2i)\) are factors of the given cubic expression. Multiplying together these two factors: \[(x-(3-2i))(x-(3+2i))=x^2-x(3-2i)-x(3+2i)+13=x^2-6x+13\] so \(x^2-6x+13\) is a quadratic factor of the cubic equation.

The remaining factor must take the form \((x+a)\) where \(a\) is real, since only one more linear factor of the cubic equation is required, and so we write: \[ x^3-7x^2+19x-13=(x^2-6x+13)(x+a).\] By inspection (consider for example the constant terms), it is clear that \(a=-1\) so that the final factor is \((x-1)\), implying that the original cubic equation has a root at \(x=1\).

Exercise

Express in terms of [value]

 \( - \sqrt{[value]} = \)
Feedback: \( - \sqrt{[value]} = - ([value]) = [value] \)

Exercise

If \(z_1 = [value] \) and \( z_2 = [value] \) determine:

1. \( z_1 + z_2 = \)
2. \( z_1 - z_2 = \)
Feedback: If \(z_1 = [value] \) and \( z_2 = [value] \) 1. \( z_1 + z_2 = [value] + [value] = [value] \) 2. \( z_1 - z_2 = [value] - [value] = [value] \)

Exercise

If \( z_1 = [value] \) and \( z_2 = [value] \) determine \( z_1 \cdot z_2. \)  Input your answer, in the form p + q*i, in the box below:

\(z_1 \cdot z_2 =\)
Feedback: We are given \( z_1 = [value] \) and \( z_2 = [value] \) \(z_1 \cdot z_2 = [value] \cdot ([value]) = [value]\)

Exercise

If \( z = [value] \) determine \( z^2. \)  Input your answer, in the form p + q*i, in the box below:

\(z^2 =\)
Feedback: We are given \(z = [value]\). Therefore \( z^2 =([value])^2 = ([value])([value]) = [value] \)

Exercise

Convert

\[\frac{[value] + [value]\,[value]}{[value] + [value]\,[value]}\]

into the form \(p+q\,[value]\), where \(p\) and \(q\) are real. Input your answer, in the form p + q*i, in the box below:

\(\displaystyle \frac{[value] + [value]\,[value]}{[value] + [value]\,[value]} =\)
Feedback: Multiplying the top and bottom by the conjugate of the denominator gives \[ \frac{\ [value] + [value]\,[value]\ }{\ [value] + [value]\,[value]\ } \times \frac{\ [value] - [value]\,[value]\ }{\ [value] - [value]\,[value]\ } = \frac{\ [value] + [value]\,[value] - [value]\,[value] - [value]\,[value]^2\ }{\ [value] + [value]\,[value] - [value]\,[value] - [value]\,[value]^2\ },\] and because \([value]^2= -1\) we obtain \[ \frac{\ [value] + [value]\,[value] - [value]\,[value] - [value]\,[value]^2\ }{\ [value] + [value]\,[value] - [value]\,[value] - [value]\,[value]^2\ } = \frac{\ [value]\ }{\ [value]\ }=[value].\]

Exercise

If \( z = [value] \) find:
1. The conjugate of \(z\), \(z^* = \)
2. The modulus of \(z\), \(|z| = \)
Feedback: 1. The conjugate of \(z = [value]\) is \(z^* = [value]\). 2. The modulus of \(z\) is \(|z| = \sqrt{[value]^2 + [value]^2 } = [value] \).

Exercise

If the two complex numbers \( z_1 = [value]+[value] \) and \( z_2 = [value]-[value]\) are equal then find the values of the real numbers \(k\) and \(l\).

\( k = \)

\( l = \)

Give your answers as rational numbers.
Feedback: Since \( z_1 = z_2 \) we can equate their real and imaginary parts. So: \[ [value]k+[value] = k+ [value] \quad \text{and} \quad [value]l-[value] = -(k-l) \] By solving these equations we find \( k = \displaystyle [value]\) and \(l = \displaystyle [value]\).

Exercise

Convert \[([value] + [value]\,[value])([value] + [value]\,[value])\] into the form \(p+q\,[value]\), where \(p\) and \(q\) are real. Input your answer, in the form p + q*i, in the box below:

\(([value] + [value]\,[value])([value] + [value]\,[value]) =\)
Feedback: Multiplying gives \[ ([value] + [value]\,[value])([value] + [value]\,[value]) = [value] + [value]\,[value] + [value]\,[value] + [value]\,[value]^2 = [value] + [value]\,[value] + [value]\,[value] - [value] = [value].\]

Exercise

Find the roots of the equation [value]. Your answers must be precise.

\(x=\)

\(x=\)
Feedback: We are given the equation \([value]\). we can then use the quadratic formula with \(a = [value]\), \(b = [value]\) and \(c = [value]\). This yields the two roots of the equation \[x = [value], \qquad x = [value].\] Remember that complex roots occur in conjugate pairs.

Exercise

Convert \[\frac{[value] + [value]\,[value]}{[value]\,[value]}\]

into the form \(p+q\,[value]\), where \(p\) and \(q\) are real. Input your answer, in the form p + q*i, in the box below:

\(\displaystyle \frac{[value] + [value]\,[value]}{[value]\,[value]} =\)
Feedback: Multiplying the top and bottom by the conjugate of the denominator gives \[ \frac{\ [value] + [value]\,[value]\ }{\ [value]\,[value]\ } \times \frac{\ - [value]\,[value]\ }{\ - [value]\,[value]\ } = \frac{- [value]\,[value] - [value]\,[value]^2\ }{\ - [value]\,[value]^2\ },\] and because \([value]^2= -1\) we obtain \[ \frac{\ [value] - [value]\,[value]\ }{\ [value] \ } = [value].\]

Exercise

Let  \(z = [value] \), where \(l\) is a real number. Find a value for \(l\) such that:
1. \(z\) is a real number, \( l = \)
2. \(z\) an imaginary number, \( l = \)
Remember that \(l\) should be a real number.
Feedback: Our complex number \( z = [value] = [value] = ([value]l+[value])+([value]-l)[value] \) 1. In order z to be real we need \( [value]-l = 0 \Rightarrow l = [value]\). 2. In order z to be imaginary we need \( [value]l+[value] = 0 \Rightarrow l = [value]\).

Exercise

Find the value of \(zz^*\) when \(z=[value]\) and \(z^*\) is its conjugate.
Feedback: We are given \(z = [value]\). Therefore its complex congujate is \(z^* = [value]\). Now expand out the brackets \( ([value])([value])=[value]\).

Exercise

If \(z=[value]\) and \(w=[value]\) find \(\displaystyle \frac{2z+3w}{2z-3w}\).
Feedback: We first find \(\displaystyle \frac{2z+3w}{2z-3w}\) by substituting \(z =[value]\) and \(w=[value]\) and simplifing. \[ \frac{2z+3w}{2z-3w} = \frac{2([value])+3([value])}{2([value])-3([value])} = \frac{[value]}{[value]}.\]; Multiplying the top and bottom by the conjugate of the denominator gives \[ \frac{[value]}{[value]} \times \frac{[value]}{[value]} = \frac{[value]}{[value]} =[value].\]

Key Point

The symbol \(i\) is such that \[i^2=-1.\] Using the normal rules of algebra it follows that \[i^3=i^2\times i=-i\qquad i^4=i^2\times i^2=(-1)\times (-1)=1\] and so on.

Simple examples of complex numbers are \(\qquad z_1=3+2i \quad\qquad z_2=-3+(2.461)i\quad\qquad z_3=17i \quad\qquad z_4=3+0i = 3\)

Generally, if \(z=a+i b\) then ‘\(a\)’ is called the real part of \(z\), or Re(\(z\)) for short, and ‘ \(b\)’ is called the imaginary part of \(z\) or Im(\(z)\) for short. Notice how both the real and imaginary parts are real numbers. The fourth example indicates how the real numbers can be considered a subset of the complex numbers.

Key Point

\[\text{If}\qquad z=a+i b\qquad\text{then}\qquad \text{Re}(z)=a\quad \text{and}\quad \text{Im}(z)=b\] Both the real and imaginary parts of a complex number are real.

Key Point

Two complex numbers \(z=a+i b\) and \(w=c+i d\), where \(a,b,c,d\) are real, are said to be equal if and only if both their real parts are the same and both their imaginary parts are the same, that is \[a=c\qquad \text{and}\qquad b=d.\]

Key Point

The modulus of a complex number \(z=a+i b\) is denoted by \(|z|\) and is defined by

\(\qquad |z|=\sqrt{a^2+b^2}\)

so that the modulus is always a non-negative real number.

Example

If \(z=3-2i\) then find \(\textrm{Re}(z)\), \(\textrm{Im}(z)\) and \(|z|\).

Solution

Here \(\textrm{Re}(z)=3\), \(\textrm{Im}(z)=-2\) and \(|z|=\sqrt{3^2+(-2)^2}=\sqrt{13}\).

Exercise

If \([value]\) is one root of the cubic equation \([value]=0\) find the two other roots. Your answers must be precise.

\(x=\)

\(x=\)
Feedback: We are given the equation \([value] = 0\). We are given that one of the roots of the polynomial is \(i\). So \(-i\) is also a root since complex roots come in conjugate pairs. Then \([value] = (x+i)(x-i)(x-a) = ([value])(x-a).\) If we expand the brackets and compare coefficients, we get \(a = [value]\). Therefore the two other roots are \(x=-i\) and \(x=[value]\).

Key Point

Complex roots to polynomial equations having real coefficients always occur in complex conjugate pairs .