# Part 2: Different Forms of Complex Numbers | Beginner’s Guide to Year 12 Ext 2 Maths

Complex numbers aren’t that complex. They just come in many different forms, and you can thank Euler and friends for that. Fortunately, this guide will break down what these forms are all about, so you can real-ise your goal (but currently imaginary) Maths marks. The study of Complex Numbers can be used to appreciate mathematics in the real world and is used in physics, engineering and statistics. In this article, we’re going to expand on our introduction to complex numbers with explanations and practice questions.

## Year 12 Extension 2 Mathematics: Different Forms of Complex Numbers

The topic Complex Numbers builds upon the existing knowledge of the real number system and involves the investigation and understanding of the imaginary numbers. This builds on prior knowledge and applications of algebra and geometry to the complex number system.

### NESA Syllabus Outcomes

NESA requires students to be proficient in the following outcomes:

#### N1.2: Geometric representation of a complex number

• Represent and use complex numbers in the complex plane
• use the fact that there exists a one-to-one correspondence between the complex number $$z=a+ib$$ and the ordered pair $$(a,b)$$
• plot the point corresponding to $$z=a+ib$$
• Represent and use complex numbers in polar or modulus-argument form, $$z=r(cosθ+isinθ)$$, where $$r$$ is the modulus of $$z$$ and $$θ$$ is the argument of $$z$$
• define and calculate the modulus of a complex number $$z=a+ib$$ as $$|z|= \sqrt{a^2+b^2}$$
• define and calculate the argument of a non-zero complex number $$z=a+ib$$ as $$arg(z)=θ$$, where $$tanθ= \frac{b}{a}$$
• define, calculate and use the principal argument $$Arg(z)$$ of a non-zero complex number $$z$$ as the unique value of the argument in the interval $$(-π,π]$$
• Prove and use the basic identities involving modulus and argument
• $$|z_1z_2|=|z_1||z_2|$$ and $$arg(z_1z_2)=arg \ z_1+arg \ z_2$$
• $$\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$$ and $$arg \left( \frac{z_1}{z_2} \right) = arg \ z_1 – arg \ z_2, \ z_2 ≠0$$
• $$|z^n|=|z|^n$$ and $$arg(z^n)=n \ arg \ z$$
• $$\left| \frac{1}{z^n} \right| = \frac{1}{|z|^n}$$ and $$arg \left( \frac{1}{z^n} \right) = -n \ arg \ z, \ z≠0$$
• $$\overline{z_1} + \overline{z_2} = \overline{z_1+z_2}$$

#### N1.3: Other representations of complex numbers

• Understand Euler’s formula, $$e^{ix}=cosx +isinx$$, for real $$x$$
• Represent and use complex numbers in exponential form, $$z=re^{iθ}$$, where $$r$$ is the modulus of $$z$$ and $$θ$$ is the argument of $$z$$
• Use Euler’s formula to link polar form and exponential form
• Convert between Cartesian, polar and exponential forms of complex numbers
• Find powers of complex numbers using exponential form
• Use multiplication, division and powers of complex numbers in polar form and interpret these geometrically
• Solve problems involving complex numbers in a variety of forms

#### N2.1: Solving equations with complex numbers

• Define and determine complex conjugate solutions of real quadratic equations
• Determine conjugate roots for polynomials with real coefficients
• Solve problems involving real polynomials with conjugate roots

## Assumed Knowledge:

Students should feel confident utilising skills like algebra, trigonometry and geometry using the real number system. Students should also be familiar with manipulating complex numbers algebraically.

## The Conjugate Theorems:

Let $$z=x+yi$$

Recall $$\overline{z}=x-yi$$

The syllabus requires the proof and use of basic identities.

• Prove $$\overline{z_1} + \overline{z_2} = \overline{z_1+z_2}$$

Let $$z_1=a+bi$$ and $$z_2=c+di$$

\begin{align*}
\overline{z_1} + \overline{z_2} &= \overline{a+bi} + \overline{c+di}\\
&= a-bi+c-di\\
&= a+c -bi-di\\
&= a+c -i(b+d)\\
&= \overline{a+c+i(b+d)}\\
&= \overline{a+bi+c+di}\\
&= \overline{z_1+z_2}\\
\end{align*}

• Prove $$\overline{z_1} \ \overline{z_2} = \overline{z_1z_2}$$

Let $$z_1=a+bi$$ and $$z_2=c+di$$

\begin{align*}
\overline{z_1} \ \overline{z_2} &= \overline{a+bi} \times \overline{c+di}\\
&= (a-bi) \times (c-di)\\
&=\overline{a(c+di)+bi(c+di)}\\
&=\overline{(a+bi)(c+di)}\\
&= \overline{z_1z_2}\\
\end{align*}

$$P(x)=ax^2+bx+c$$

If $$α$$ is a root of the quadratic equation, and $$α$$ is a non-real root, then $$\overline{α}$$ must also be a root

If $$α$$ and $$β$$ are roots of $$P(x)=ax^2+bx+c$$, then $$α+β= \ – \frac{b}{a}$$ and $$αβ= \frac{c}{a}$$

So $$P(x)=ax^2+bx+c$$ can also be rewritten as

 $$P(x)=x^2-(Sum \ of \ Roots)x+(Product \ of \ Roots)=0$$

If this equation has non-real roots then

\begin{align*}
P(x) &=x^2-(α+\overline{α})x+α\overline{α} =0\\
P(x) &=x^2-2Re(α)x+α \overline{α}=0\\
\end{align*}

#### Example 1:

If one zero of a real, monic quadratic equation is $$4-i$$

a) State the other root

b) Find the equation of the quadratic equation

#### Solution:

a) $$4+i$$ (by the conjugate root theorem)

b) $$x^2-8x+17$$

## Cubic Roots:

Cubic Equations have the general form:

$$P(x)=ax^3+bx^2+cx+d$$

Given a cubic equation has real coefficients, then there will be at least one real root. From this, we can break it down into 3 different conditions:

• One real zero (and two imaginary zeros)
• Two real zeros (and one imaginary zero – can this occur?)
This will not occur as complex roots occur in conjugate pairs.
• Three real zeros

## The Argand Diagram:

Complex Numbers can be expressed graphically on an Argand Diagram where $$z=x+yi$$ is equal to the coordinates $$P(x,y)$$.

This is similar to the $$xy$$ plane, except real numbers are represented on the $$x$$ axis whilst imaginary numbers are represented on the $$y$$ axis.

Recall the conversion of degrees to radians

\begin{align*}
180 \times \frac{x}{π} \ degrees &= π \times \frac{x}{π} \ radians\\
180 \times \frac{x}{π} \ degrees &= x \ radians\\
\end{align*}

### Example 1:

Graph the following complex numbers on an Argand diagram
a) $$z_1=4i$$
b) $$z_2=-2-2i$$
c) $$z_3=5-3i$$

### Solution: ## The Modulus-Argument of a Complex Number

When represented on an Argand Diagram, a complex number can also be expressed as its angle and distance from the origin.

The length or modulus of $$z$$ is written as $$|z|$$

$$|z|=|x+yi|=\sqrt{x^2+y^2}$$

The angle or argument is defined from the positive, real axis and is rotated anti-clockwise.

The argument can be solved by $$tanθ= \frac{y}{x}$$, and the sign is determined by the quadrant in which the point lies in. The Principal Argument: in which $$θ$$ has a unique value lies here:

Principal argument: $$-π<Arg(z)≤π$$

As observed from the argand diagram where $$z=x+yi$$

\begin{align*}
x &=|z|cosθ\\
y &=|z|sinθ\\
\end{align*}

If $$P(x,y)$$ lies on a circle with radius $$r$$, then $$|z|=r$$

\begin{align*}
x &=|z|cosθ\\
&=rcosθ\\
y &=|z|sinθ\\
&=rsinθ\\
\\
cosθ+isinθ&=cisθ\\
\end{align*}

Therefore $$z=rcisθ$$

### Example 1:

Given $$z_1=2+2i$$ and $$z_2= \sqrt{3}+i$$. Express following complex numbers in Mod-Arg form

a) $$z_1$$

b) $$z_2$$

c) $$\frac{z_1}{z_2}$$

d) $$z_1z_2$$

### Solution:

a)

\begin{align*}
z_1 &= 2+2i=2(1+i)\\
|z_1|&= \sqrt{1+1}=\sqrt{2}\\
arg(z_1) &= tan^{-1} \left( \frac{1}{1} \right) = \frac{π}{4}\\
z_1 &= 2\sqrt2 \ cis \frac{π}{4}\\
\end{align*}

b)

\begin{align*}
z_2 &= \sqrt3 +i\\
|z_2|&= \sqrt{3+1}=2\\
arg(z_2) &= tan^{-1} \left( \frac{1}{\sqrt3} \right) = \frac{π}{6}\\
z_2 &= 2 \ cis \frac{π}{6}\\
\end{align*}

c)

$$\frac{z_1}{z_2} = \frac{2\sqrt{2} \ cis \frac{π}{4}}{2 \ cis \frac{π}{6}} = \sqrt{2} \ cis \ \frac{π}{12}$$

d)

$$z_1z_2 = 2 \sqrt2 \ cis \frac{π}{4} \times 2 \ cis \frac{π}{6} = 4 \sqrt2 \ cis \ \frac{5π}{12}$$

### Important properties to remember:

\begin{align*}
|z_1z_2| &= |z_1| \times |z_2|\\
arg(z_1z_2) &= arg(z_1)+arg(z_2)\\
\left| \frac{z_1}{z_2} \right| &= \frac{|z_1|}{|z_2|}\\
arg \left( \frac{z_1}{z_2} \right) &= arg(z_1) -arg(z_2)\\
\end{align*}

### Example 2:

Simplify the following expressions in Mod-Arg form

a) $$4cis \left( \frac{5π}{12} \right) \times 3cis \left( \frac{π}{4} \right)$$

b) $$\frac{3cis \left( \frac{π}{2} \right)}{4cis \left( – \frac{π}{6} \right)}$$

### Solution:

a) $$12cis \frac{2π}{3}$$

b) $$\frac{3}{4}cis\left( \frac{2π}{3} \right)$$

When complex numbers are raised to an integer $$n$$ power where $$n$$ is a positive integer:

\begin{align*}
|z^n| &= |z|^n\\
arg(z^n) &= n  \ arg(z)\\
\end{align*}

This property forms the foundation of the De Moivre’s Theorem

### Example 3:

Simplify the following expressions in Mod-Arg form

a) $$(4cis \left( \frac{π}{3} \right))^3$$

b) $$(2cis \left( – \frac{2π}{5} \right))^3$$

### Solution:

a) $$64cis(π)=64(-1+0i)=-64$$

b) $$8cis \left(- \frac{6π}{5} \right) = 8cis \left( \frac{4π}{5} \right)$$

## The Exponential Form of a Complex Number

\begin{align*}
cosθ+isinθ &= e^{iθ}\\
r(cosθ+isinθ) &= re^{iθ}\\
\end{align*}

As the argument in a complex number is periodic, and repeats every $$2π$$, then

$$e^{iθ}=e^{i(θ+2kπ)}$$ where $$k$$ is an integer

Summary: the different forms of expressing a complex number $$z$$

1. $$x+yi$$ (Cartesian)

2. $$|z|(cosθ+isinθ)$$ (Mod-Arg)

3. $$rcisθ$$ (Mod-Arg)

4. $$re^{iθ}$$ (Exponential)

### Example 1:

Express the following in exponential form

a) $$\sqrt3 -i$$

b) $$-2-2i$$

### Solution:

a)

\begin{align*}
|z|&=2, arg(z)= \ – \frac{π}{6}\\
z &= 2e^{- \frac{iπ}{6}}\\
\end{align*}

b)

\begin{align*}
|z|&=2\sqrt2, arg(z)= \ – \frac{3π}{4}\\
z &= 2\sqrt2e^{- \frac{i3π}{4}}\\
\end{align*}

## Concept Check Questions

1. Have a go at proving the rest of the basic complex identities

a) $$z \bar{z} = |z|^2$$

b) $$z + \bar{z} = 2Re(z)$$

c) $$z- \bar{z} = 2iIm(z)$$

2. On the complex plane, $$P$$ represents the complex number $$z=x+yi$$, where $$|z|=1$$ and $$\frac{π}{4}<θ<\frac{π}{2}$$. Graph $$\bar{z}$$, $$2z$$, $$iz$$ and $$\frac{1}{z}$$

3. Express the following in the Cartesian form

a) $$cis \left( \frac{5π}{6} \right)$$

b) $$(1+\sqrt3i)^{10}$$

c) $$(2e^{\frac{iπ}{6}})^{15}$$

## Concept Check Solutions

1. Proof

2. 3.

a) $$cos \left( \frac{5π}{6} \right) +isin \left( \frac{5π}{6} \right) = \ – \frac{\sqrt3}{2}+ \frac{1}{2}i$$

b)

\begin{align*}
&(2cis \frac{π}{3})^{10}\\
&= 1024cis \left( \frac{10π}{3} \right)\\
&= 1024cis \left( -\frac{2π}{3} \right)\\
&= 1024 \left( -\frac{1}{2} – \frac{\sqrt{3}}{2}i \right)\\
&= -512 -512\sqrt3i\\
\end{align*}

c)

\begin{align*}
&(2e^{\frac{iπ}{6}})^{15}\\
&= 2^{20}e^{\frac{i15π}{6}}\\
&= 2^{20}e^{\frac{i5π}{2}}\\
&= 2^{20}e^{\frac{iπ}{2}}\\
&= 2^{20}i\\
\end{align*}

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