Complex numbers are used in mathematics and physics to model and represent a variety of phenomena that cannot be explained with real numbers alone.  Complex numbers are a combination of real and imaginary numbers. This chapter discusses:

• Imaginary numbers
• Real and imaginary numbers which form complex numbers
• Complex number manipulation
• Complex conjugates
• Argand diagram
• Polar form of a complex number
• Argument of a complex number
• Manipulation of complex numbers in Polar form
• DeMoivres Theorem

Do you need this chapter?

Here are some questions on the topics in this chapter to consider. If you can do these you may skip this chapter.

If you need to review this topic continue reading.

Imaginary numbers

Equations such as [latex]x+1=5[/latex], [latex]2x=8[/latex] and [latex]x^{2}-16=0[/latex] can all be solved within the real number system, the answer being [latex]4[/latex] in all the previous cases. But there is no real number which satisfies [latex]x^{2}+1=0[/latex] as this leads to an answer of [latex]x=\sqrt{-1}[/latex]  which is not possible using only the real number system.

To obtain solutions to this and other similar equations, a new branch of maths was developed called “complex numbers”, which incorporates real and imaginary numbers.

We define an imaginary number [latex]i[/latex] such that [latex]i^{2}=-1[/latex].

Which leads to the imaginary number  [latex]i=\sqrt{-1}.[/latex]

Note: In electrical engineering, it is common to use [latex]j[/latex] instead of [latex]i[/latex].

Using [latex]i[/latex] , it is noted that:

[latex]i^{2} = -1.[/latex]

[latex]i^{3} = i^{2}i = (-1)i = -i.[/latex]

[latex]i^{4} = i^{2}i^{2} = (-1)(-1) = 1.[/latex]

[latex]i^{5} = i^{2}i^{2}i = (-1)(-1)i = i.[/latex]

[latex]i^{6} = i^{2}i^{2}i^{2} = (-1)(-1)(-1) = -1.[/latex]

etc.

Complex numbers

Complex numbers are a combination of real numbers and imaginary numbers such as [latex]3+4i[/latex], [latex]4+0i[/latex] or [latex]4[/latex], [latex]0+5i[/latex] or [latex]5i[/latex]. The real and imaginary numbers are a subset of complex numbers. We can define [latex]z[/latex] such that [latex]z = 3+4i[/latex] so every instance of [latex]z[/latex] now represents [latex]3+4i[/latex]. We may also define [latex]w= 7+3i[/latex] so that every instance of [latex]w[/latex] now represents [latex]7+3i[/latex]. This is described as the rectangular form of a complex number (sometimes also known as cartesian form).

A number [latex]z[/latex] of the form [latex]z=x+yi[/latex] where [latex]x[/latex] and [latex]y[/latex] are both  real numbers is called a “complex number”.

• [latex]x[/latex] is called the real part of [latex]z[/latex] denoted by [latex]Re\ z[/latex]    [latex]\\[/latex]
• [latex]y[/latex] is called the imaginary part of [latex]z[/latex] denoted by [latex]Im\ z[/latex]

A complex number is represented usually by the letter [latex]z[/latex], however, any letter can represent a complex number. i.e. [latex]z = 3 + 5i[/latex] or [latex]w=4-3i[/latex].

Two complex numbers are equal when both the real components and the imaginary components are equal such as :

[latex]3+4i =3+ \left(2\right)^{2}i=3+\sqrt{16} \ i[/latex] or

[latex]x+yi=2-3i[/latex] is only true if [latex]x=2[/latex] and [latex]y=-3[/latex].

Now try the exercises below.

Addition and subtraction of complex numbers

To add or subtract complex numbers we add or subtract the real and imaginary parts separately:

\begin{align*}\left(a + bi \right)  + \left(c + di\right)  &= \left(a + c\right) + \left(b + d\right)i  \end{align*}

and

\begin{align*}\left(a+bi\right) -\left(c+di\right) &=\left(a- c\right) +\left(b- d\right)i . \end{align*}

Now try the exercise below.

Multiplication of complex numbers

To multiply complex numbers, use the formulae:

\begin{align*}k(a+bi)=ka+kbi\end{align*}

and

\begin{align*}(a+bi)(c+di)=(ac+bdi^{2})+(ad+bc)i=(ac-bd)+(ad+bc)i,\end{align*}

where [latex]k[/latex], [latex]a[/latex], [latex]b[/latex], [latex]c[/latex], and [latex]d[/latex] are real numbers.

Remember that  [latex]i^2=-1[/latex].

Now try the exercise below.

Complex conjugates

A pair of complex numbers of the form  [latex]a+bi[/latex]  and  [latex]a-bi[/latex]  are called complex conjugates.

If  [latex]z=x+yi[/latex]  then the conjugate of [latex]z[/latex] is denoted by  [latex]\overline{z}=x-yi.[/latex]

Some properties of conjugates

1. The product of a conjugate pair of complex numbers is a real number.

\begin{align*} z\overline{z} & = (x+yi)(x-yi) \\ & =x^{2}-xyi+xyi-y^{2}i^{2} \\ & = x^{2}-y^{2}i^{2}\\ & = x^{2}+y^{2}. \end{align*}

2. If [latex]z_{1}[/latex] and [latex]z_{2}[/latex] represent two conjugate numbers then:

\begin{align*} \overline{z_{1}+z_{2}} & = \overline{z_1}+ \overline{z_2} . \end{align*}

3. If [latex]z_{1}[/latex] and [latex]z_{2}[/latex] represent two conjugate numbers then:

\begin{align*} \overline{z_{1} \times z_{2}} & = \overline{z_1} \times \overline{z_2} . \end{align*}

Now complete the exercise below.

Division of complex numbers

if [latex]z_{1}=a+bi[/latex] and [latex]z_{2}=c+di[/latex] are two complex numbers then

To express [latex]\dfrac{z_{1}}{z_{2}}[/latex]  in the form [latex]x+yi[/latex] we make use of the conjugate to “rationalise” the denominator (make the denominator a real number by multiplying the numerator and the denominator by the conjugate of the original denominator).

\begin{align*} \frac{z_{1}}{z_{2}} &= \frac{a+bi}{c+di} \\ & = \frac{(a+bi)}{(c+di)}\times  \frac{(c-di)}{(c-di)} \\ & = \frac{ac-adi+cbi-bdi^2}{c^2-d^2i^2} \\ & = \frac{(ac+bd)+(cb-ad)i}{c^2+d^2} .\end{align*}

Always remember that   [latex]i^2=-1[/latex].

Now try the exercises below:

Argand diagram

An Argand Diagram is a geometrical representation of the set of complex numbers. The complex number[latex]z=x+yi[/latex] can be plotted as a point represented by the ordered pair [latex](x,y)[/latex] on the complex number plane. Note that the complex plane has the horizontal axis labelled as real [latex]\left(Re \right)[/latex] and the vertical axis labelled as imaginary [latex]\left(Im \right)[/latex]. When a complex number is expressed in the form [latex]z=x+yi[/latex] it is said to be in rectangular or cartesian form. Figure 15.1 shows the plot of the following complex numbers:

• [latex]4+3i[/latex]
• [latex]-2+i[/latex]
• [latex]-5-4i[/latex]
• [latex]3-2i.[/latex]

Polar form of a complex number

Remember that when a complex number is expressed in the form [latex]z=x+yi[/latex] , it is said to be in rectangular or Cartesian form. But a point [latex]P[/latex] with Cartesian coordinates [latex](x,y)[/latex] can also be represented by the polar coordinates [latex]\left(r,\theta\right)[/latex] where [latex]r[/latex] is the distance of the point [latex]P[/latex] from the origin and [latex]\theta[/latex] is the angle that [latex]\overline{OP}[/latex] makes with the positive x-axis as shown below:

The distance [latex]r[/latex] is called the modulus of the complex number and is denoted by [latex]\left|z\right|.[/latex] Using Pythagoras's Theorem, we can see from the figure above that:

\begin{align*} r&=\left|z\right|\\&=\left|x+yi\right|\\&=\sqrt{x^{2}+y^{2}.} \end{align*}

The angle [latex]\theta[/latex] is called the argument of the complex number. Note that the argument is NOT unique because we can add any multiple of [latex]2\pi[/latex] to it and get the same result. We discuss this further in the Quadrants section below.

Looking at Fig. 15.2  we notice that:

\begin{align*} x&=r\cos\theta\\y&=r\sin\theta. \end{align*}

Using the relationships above, a complex number [latex]z=x+yi[/latex] may be written as follows:

\begin{align*}
z&=x+yi\\
&=r\cos\theta+r\sin\theta i\\
&=r\left(\cos\theta+i\sin\theta\right)\\
&=r\, \textrm{cis} \ \theta
\end{align*}

This is called polar form. Note that we usually use the abbreviation [latex]{\textrm{cis}{\ \theta=\cos\theta+i\sin\theta}}[/latex]. So a complex number in polar form is written as [latex]z=r\ {\textrm{cis}{\,\theta}}[/latex] where:

\begin{align*} r&=\sqrt{x^{2}+y^{2}}\\\sin\theta&=\frac{y}{r}\\\cos\theta&=\frac{x}{r}\\\tan\theta&=\frac{y}{x} \ \textrm{ in quadrant one.}  \end{align*}

Which leads to the important concept:  \begin{aligned} \theta&={\tan}^{-1}\left(\frac{y}{x}\right) \ \textrm{ in quadrant one.}  \end{aligned}

All of these can be derived from the figure above. Also remember the Pythagorean theorem [latex]x^2+y^2=r^2[/latex].

Quadrants

The coordinate plane and the complex plane is divided into an upper and lower section by the [latex]x[/latex]-axis. It is further divided into quadrants by the [latex]y[/latex]-axis. These four quadrants are numbered from one to four in an anti-clockwise direction as shown below:

Remember a full circle of [latex]360°[/latex] is equivalent to [latex]2\pi[/latex] radians.  So if you have an angle of [latex]2\pi[/latex] it is equivalent to an angle of [latex]4\pi[/latex] because you have rotated to the same position as an angle of [latex]2\pi[/latex]. The angle  [latex]\pi[/latex] radian is equivalent to [latex]3\pi[/latex] radian because you end up in the same location. ie [latex]\pi[/latex] is equivalent to [latex]\pi+2\pi = 3\pi[/latex].

The Argument of a complex number

The angle [latex]\theta[/latex] is called the argument of the complex number. In the previous section, we mentioned that the argument of a complex number is not unique because any multiple of [latex]2\pi[/latex] can be added to the argument and results in the same number.  Zero radians is the same as [latex]2\pi[/latex] radians and [latex]\pi/4[/latex] is the same as [latex](\pi/4+2\pi=9\pi/4)[/latex] radians. Notice that you can add or subtract [latex]2\pi[/latex] and obtain the same complex number:

\begin{align*} r \text{cis}\,\theta &=r \text{cis}\left(\theta+2\pi k\right),\;k\in\mathbb{Z}. \end{align*}

Where the symbol [latex]\mathbb{Z}[/latex] denotes the set of integers (whole numbers and zero),

\begin{align*}  \mathbb{Z}&=\left\{ 0,\pm1,\pm2,\ldots\right\} .  \end{align*}

To get a unique argument, we introduce the function [latex]Arg\left(z\right)[/latex] where [latex]Arg\left(z\right)[/latex] is the single value of the argument of the complex number [latex]z[/latex] that lies in the interval [latex]\left(-\pi,\pi\right][/latex]. That is, \begin{aligned}\theta = Arg(z) \text{ when: }  \left(-\pi<Arg\left(z\right) \leq \pi\right]  \end{aligned}

How to express a complex number in polar form

Given a complex number [latex]z=x+yi,[/latex] we want to convert to polar form. The steps are:

1. Plot the point on an Argand diagram.

2. Calculate [latex]r=\sqrt{x^{2}+y^{2}}[/latex].

3. Determine the argument of [latex]z[/latex] using trigonometric relations and the plot in step 1. This step may require you to know some exact values of [latex]\sin, \cos[/latex] and [latex]\tan[/latex] in different quadrants.

4. Determine  [latex]\theta=Arg\left(z\right)[/latex]. That is, make sure [latex]-\pi \,[/latex][latex]\lt[/latex][latex]\,  \theta\leq\pi[/latex].

5. Write [latex]z=r\,\text{cis}\,\theta[/latex].

Remember: [latex]\text{cis}\,\theta = \cos \theta + i \sin \theta .[/latex]

Examples: Expressing complex numbers in polar form

Example 1: Write the complex number [latex]z=1+i[/latex] in polar form.

Solution:

The point of interest is plotted below:

 

We have [latex]x=1[/latex] and [latex]y=1[/latex]. So

\begin{aligned} r&=\sqrt{x^{2}+y^{2}}\\&=\sqrt{1^{2}+1^{2}}\\&=\sqrt{2}. \end{aligned}

Referring to the figure above, we can see that:

\begin{aligned}\tan\theta &=\frac{y}{x}\\&=\frac{1}{1}\\&= 1.\end{aligned}

so from the figure above we note that [latex]1+i[/latex] is in the first quadrant.

\begin{aligned}\theta &=\tan^{-1}\left(1\right) \\&=\frac{\pi}{4}.\end{aligned}

Noting [latex]-\pi\lt \dfrac{\pi}{4}\le\pi[/latex] , we have [latex]Arg(z)=Arg(1+i)=\dfrac{\pi}{4}.[/latex]

So in Polar Form:

\begin{aligned}z=\sqrt{2} \,\text{cis}\left(\dfrac{\pi}{4}\right). \end{aligned}

Example 2:

Write the complex number [latex]z=-1-\sqrt{3}i[/latex] in polar form.

Solution:

The point of interest is plotted below

We have [latex]\ x=-1[/latex] and [latex]y=-\sqrt{3}[/latex].So:\begin{aligned}r&=\sqrt{x^{2}+y^{2}}\\&=\sqrt{\left(-1\right)^{2}+\left(\sqrt{3}\right)^{2}}\\&=\sqrt{4}\\&=2.\end{aligned}Referring to Fig. 15.5, we see that:\begin{aligned}\tan\theta&=\frac{y}{x}\\&=\frac{-\sqrt{3}}{-1}\\&=\sqrt{3}. \end{aligned}So:\begin{aligned}\theta&=\tan^{-1}\left(\sqrt{3}\right)\\&=\dfrac{\pi}{3}.\end{aligned}Possible values for the argument of [latex]z[/latex] shown from the [latex]0[/latex] position are:

1.  [latex]\ \pi+\theta=\pi+\dfrac{\pi}{3}=\dfrac{4\pi}{3}.[/latex]

2.  [latex]\ -\pi+\theta=-\pi+\dfrac{\pi}{3}=-\dfrac{2\pi}{3}.[/latex]

We choose [latex]Arg\left(z\right)=-\dfrac{2\pi}{3}[/latex] as it lies in the interval [latex]-\pi \lt   \theta\leq\pi[/latex].

Hence:

\begin{aligned} z=2 \text{cis}{\left(-\dfrac{2\pi}{3}\right)}.\end{aligned}

Example 3:

Express [latex]2\text{cis}\left(\dfrac{4\pi}{3}\right)[/latex] in rectangular form [latex]x+yi[/latex].

Solution:

\begin{aligned}2\text{cis}\left(\dfrac{4\pi}{3}\right)&=2\left[\cos\left(\frac{4\pi}{3}\right)+i\sin\left(\frac{4\pi}{3}\right)\right] \\  &=2\left[-\frac{1}{2}+\left(-\frac{\sqrt{3}}{2}\right)i\right]\\  &=-1-\sqrt{3}i.  \end{aligned}

 

Now try the exercises below:

Addition and subtraction of complex numbers in polar form.

Complex numbers in polar form are best converted to the form [latex]x+yi[/latex] before addition or subtraction. To add (or subtract) two complex numbers, you add (or subtract) the real parts and the imaginary parts separately.

Example: Addition and subtraction of complex numbers in polar form 

1. If [latex]z_{1}=\sqrt{2}\textrm{cis}\left(-\dfrac{\pi}{4}\right)[/latex] and [latex]z_{2}=2\textrm{cis}\left(\dfrac{4\pi}{3}\right)[/latex] , find [latex]z_{1}+z_{2}[/latex].

Solution:

Figure 15.6 shows [latex]z_{1}.[/latex]

\begin{aligned}z_{1}&=\sqrt{2}\text{cis}\left(\frac{-\pi}{4}\right)\\&=\sqrt{2}\left[\cos\left(\frac{-\pi}{4}\right)+\sin\left(\frac{-\pi}{4}\right)i\right] \\  &=\sqrt{2}\left[\frac{1}{\sqrt{2}}+\left(-\frac{1}{\sqrt{2}}\right)i\right]\\  &=1-i.  \end{aligned}Figure 15.7 shows [latex]z_{2}.[/latex] 

\begin{aligned}z_{2}&=2\text{cis}\left(\frac{4\pi}{3}\right)\\&=2\left[\cos\left(\frac{4\pi}{3}\right)+\sin\left(\frac{4\pi}{3}\right)i\right] \\  &=2\left[-\frac{1}{2}+\left(-\frac{\sqrt{3}}{2}\right)i\right]\\  &=-1-\sqrt{3}i.  \end{aligned}Hence\begin{aligned}z_{1}+z_{2}&= \left(1-i\right)+\left(-1-\sqrt{3}i\right) \\&=1-1-i-\sqrt{3}i \\& = -\left( 1+\sqrt{3} \right)i .\end{aligned}Note: the answer could then be converted back into polar form, if this is required.

Multiplication and division of complex numbers in polar form

If [latex]z_{1}=r_{1} \, cis \, \theta_{1}[/latex] and [latex]z_{2}=r_{2} \, cis \, \theta_{2}[/latex] then it can be shown using trigonometric identities that:

\begin{aligned}  z_{1}z_{2} &=\left(r_{1} \,cis \theta_{1}\right)  \left(r_{2}\, cis \,  \theta_{2}\right) \\ \\ &=r_{1}r_{2}\, cis \, \left(\theta_{1}+\theta_{2}\right)   \end{aligned}

and

\begin{aligned}  \frac{z_{1}}{z_{2}}  &= \frac{r_{1}\, cis \theta_{1}}{r_{2}\,cis \theta_{2}} \\ \\ &= \frac{r_{1}}{r_{2}} \, cis \left(\theta_{1}-\theta_{2}\right). \end{aligned}

Examples: Multiplication and division in polar form

1. If [latex]z_{1}=2 \, \text{cis} \, \dfrac{\pi}{4}[/latex] and [latex]z_{2}=3 \, \text{cis} \dfrac{5\pi}{6}[/latex] find  [latex]z_{1}z_{2}[/latex]  in polar form, [latex]-\pi\lt \theta \leq \pi[/latex] .

Solution:

\begin{aligned} z_{1}z_{2}&=(2\,\textrm{cis}\frac{\pi}{4})\times(3\,\textrm{cis}\frac{5\pi}{6}) \\ &=6\,\textrm{cis}(\frac{\pi}{4}+\frac{5\pi}{6})   \\ &=6\,\textrm{cis}\left(\frac{13\pi}{12}\right) \\ &=6\,\textrm{cis}\left(-\frac{11\pi}{12}\right)\ \ \text{ since }\ -\pi\lt\theta\leq\pi.\end{aligned}

Note that [latex]\dfrac{13\pi}{12}=-\pi+\dfrac{1\pi}{12}=-\dfrac{11\pi}{12} \ \text{ since }\ -\pi\lt\theta\leq\pi.[/latex]

 

2. If [latex]u=1+3i[/latex] and [latex]v=2-i[/latex] find [latex]\dfrac{u}{v}[/latex] in polar form with [latex]-\pi\lt\theta\leq\pi[/latex] .

There are two possible approaches to this problem. The first converts [latex]u[/latex] and [latex]v[/latex] to polar form and the second uses the rectangular form and then converts to polar form. We show both solutions below.

Solution 1:

We first plot the points [latex]u[/latex] and [latex]v[/latex]:

For [latex]\ u=1+3i[/latex] , [latex]x=1[/latex] and [latex]y=3.[/latex]

So  [latex]\ r=\sqrt{1^{2}+3^{2}}= \sqrt{10}.[/latex]

From the figure,  [latex]\theta=\tan^{-1}\left(\dfrac{3}{1}\right)=1.25[/latex] radians.

Therefore [latex]u=\sqrt{10} \, \text{cis} \left ( 1.25 \right).[/latex]

For [latex]\ v=2-i[/latex] ,  [latex]x=2[/latex] and [latex]y=-1[/latex].

So [latex]r=\sqrt{2^{2}+(-1)^{2}}= \sqrt{5}.[/latex]

From the figure, [latex]\ \phi=\tan^{-1}\left(-\dfrac{1}{2}\right)=-0.46[/latex] radians

Therefore [latex]v=\sqrt{5}\,\textrm{cis}(-0.46).[/latex]

Then

\begin{aligned} \frac{u}{v}&=\frac{\sqrt{10}\,\textrm{cis}\left(1.25\right)}{\sqrt{5}\,\textrm{cis}(-0.46)} \\ &=\frac{\sqrt{10}}{\sqrt{5}}\,\textrm{cis}(1.25-(-0.46)) \\ &=\frac{\sqrt{2}\sqrt{5}}{\sqrt{5}}\,\textrm{cis}(1.25+0.46) \\  &=\sqrt{2}\,\textrm{cis}(1.71). \end{aligned}

 

Solution 2:

We use the conjugate to rationalise the denominator. The conjugate of [latex]2-i[/latex] is [latex]2+i[/latex] so:

\begin{aligned}  \frac{u}{v}&=\frac{1+3i}{2-i} \\ &=\frac{1+3i}{2-i}\times\frac{2+i}{2+i} \\ &=\frac{2+i+6i+3i^{2}}{4+1} \\ &=\frac{2+i+6i-3}{5} \\ &=-\frac{1}{5}+\frac{7}{5}i .  \end{aligned}

Therefore [latex]x=-\dfrac{1}{5}=0.2[/latex] and [latex]\ y=\dfrac{7}{5}=1.4.[/latex] .

We can now calculate [latex]r[/latex] and [latex]\theta[/latex] so that we can express the answer in polar form. The point [latex]-\dfrac{1}{5}+\dfrac{7}{5}i[/latex] is plotted below.

 

[latex]r=\sqrt{x^{2}+y^{2}}=\sqrt{(-0.2)^{2}+(1.4)^{2}}=\sqrt{2}.[/latex][latex]\tan\theta=\dfrac{0.2}{1.4}=0.143 \ \implies \theta=0.142[/latex] radians. So that (see diagram)  [latex]Arg\left(-\dfrac{1}{5}+\frac{7}{5}i\right)=0.142+\dfrac{\pi}{2}=1.713[/latex]. Hence:\begin{aligned}  \frac{u}{v} &=r \,\textrm{cis} \theta \\  &=\sqrt{2} \,\textrm{cis}(1.71),  \end{aligned}as in solution 1.

Try the exercises below.

Powers of complex numbers using DeMoivres theorem

Recall that a complex number

\begin{aligned} z&=x+iy \end{aligned}

may be written in polar form as

\begin{aligned} z&=r\,\text{cis}\theta \end{aligned}

where

\begin{aligned} r=\sqrt{x^{2}+y^{2}} \end{aligned}

and

\begin{aligned} \theta = Arg\left(z\right)\ \ where -\pi\lt\theta\le\pi .\end{aligned}

De Moivre's Theorem states that [latex]\begin{aligned}  \text{ if}\ z&=r\,\text{cis}\left(\theta\right)   \text{ then}\ \ z^{n}=r^{n}\,\text{cis}\left(n\theta\right). \end{aligned}[/latex]

Now complete the exercises on De Moivres theorem.

Roots of a complex number

The [latex]n[/latex]th roots of [latex]z=r \, \text{cis}\,\theta[/latex] where [latex]z\neq0[/latex] are given by:

[latex]\sqrt[n]{r\, \text{cis}\,\theta} =\sqrt[n]{r}\, \text{cis}\left(\dfrac{\theta}{n}+\dfrac{2\pi k}{n}\right) \  \ \ \textrm{ for } k=0,1,2,.....,n-1[/latex]

or written alternatively as

[latex]z^{\frac{1}{n}} =r^{\frac{1}{n}}\, \text{cis}\left(\dfrac{\theta+2\pi k}{n}\right)\  \ \ \textrm{ for } k=0,1,2,.....,n-1.[/latex]

There are [latex]n[/latex] roots of the complex number [latex]z^{\frac{1}{n}}.[/latex]

 

Note that the roots lie on a circle of radius [latex]2^{0.25}[/latex] and are equally spaced with an angle of [latex]\dfrac{6\pi}{12}=\dfrac{\pi}{2}[/latex] between them. This is true in general. The [latex]n^{\text{th}}[/latex] roots of a complex number will lie on a circle of radius [latex]r^n[/latex] with a spacing of [latex]\dfrac{2\pi}{n}[/latex] between them. This fact may be used to find roots given one root, or to check calculated roots.

Example: Roots of a complex number II

1. If [latex]z=2\,\text{cis}\left(\dfrac{\pi}{2}\right)[/latex] is a root of [latex]z^3=-8i[/latex], find the other roots.

Solution:

There will be three roots and each will be separated by [latex]\dfrac{2\pi}{3}[/latex]. The roots are

\begin{align*}
z&=2\,\text{cis}\left( \frac{\pi}{2}+\frac{2k\pi}{3}\right)
\end{align*}

where [latex]k=0\,,1\,,2[/latex].

For [latex]k=0[/latex] we get the given root [latex]z=2\,\text{cis}\left(\dfrac{\pi}{2}\right)[/latex].

For k=1 we get

\begin{align*}
z&=2\,\text{cis}\left( \frac{\pi}{2}+\frac{2\pi}{3}\right) \\
&=2\,\text{cis}\left( \frac{3\pi}{6}+\frac{4\pi}{6}\right) \\
&=2\,\text{cis}\left( \frac{7\pi}{6}\right).
\end{align*}

The angle [latex]\dfrac{7\pi}{6}[/latex] is out of the range [latex]-\pi \lt \theta\leq\pi[/latex] but is the same as [latex]-\dfrac{5\pi}{6}[/latex]. Hence the root is [latex]z=2\,\text{cis}\left(- \dfrac{5\pi}{6}\right)[/latex].

For k=2 we get

\begin{align*}
z&=2\,\text{cis}\left( \frac{\pi}{2}+\frac{4\pi}{3}\right) \\
&=2\,\text{cis}\left( \frac{3\pi}{6}+\frac{8\pi}{6}\right) \\
&=2\,\text{cis}\left( \frac{11\pi}{6}\right).
\end{align*}

Again, the angle [latex]\dfrac{11\pi}{6}[/latex] is out of range but is equivalent to [latex]- \dfrac{\pi}{6}[/latex] hence the root is [latex]z= 2\,\text{cis}\left(- \dfrac{\pi}{6}\right)[/latex].

The additional two roots are

\begin{align*}
z&=2\,\text{cis}\left( -\frac{5\pi}{6}\right) \\
z&= 2\,\text{cis}\left(- \frac{\pi}{6}\right).
\end{align*}

 

Try the following exercises.

Exercises: Roots of complex numbers

1.  Solve, giving the answers in polar form with [latex]-\pi\lt\theta\leq\pi[/latex].

\begin{aligned}(a)\  & z^{3}=-1 & (b)\  & z^{4}=16i & (c)\  & z^{3}=\sqrt{6}-\sqrt{2}i\,. \end{aligned}

2. If [latex]\sqrt{3}+i[/latex] is one solution of [latex]z^{3}=8i[/latex], use a diagram to find the other solutions in rectangular form.

 

Answers

 

1. (a) [latex]\,\text{cis}\left(\dfrac{\pi}{3}\right)\ , \,\text{cis}\left(\pi\right)\ , \,\text{cis}\left(-\dfrac{\pi}{3}\right)[/latex]

1. (b) [latex]2 \,\text{cis}\left(\dfrac{\pi}{8}\right)\ , \,\text{cis}\left(\dfrac{5\pi}{8}\right)\ , \,\text{cis}\left(\dfrac{-7\pi}{8}\right)\ , \,\text{cis}\left(\dfrac{-3\pi}{8}\right)[/latex]

1. (c) [latex]\sqrt{2} \,\text{cis}\left(-\dfrac{\pi}{18}\right)\ ,\ \sqrt{2} \,\text{cis}\left(\dfrac{11\pi}{18}\right)\ ,\ \sqrt{2} \,\text{cis}\left(-\dfrac{13\pi}{18}\right)[/latex]

2. [latex]-\sqrt{3}+i[/latex] and [latex]-2i.[/latex]