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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.
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 requires students to be proficient in the following outcomes:
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.
Let \(z=x+yi\)
Recall \( \overline{z}=x-yi\)
The syllabus requires the proof and use of basic identities.
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*}
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)\\
&= ac-adi-bci-bd\\
&=ac-bd-i(ad+bc)\\
&=\overline{ac-bd+i(ad+bc)}\\
&=\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*}
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
a) \(4+i\) (by the conjugate root theorem)
b) \(x^2-8x+17\)
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:
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° &= π \ radians\\
180 \times \frac{x}{π} \ degrees &= π \times \frac{x}{π} \ radians\\
180 \times \frac{x}{π} \ degrees &= x \ radians\\
\end{align*}
Graph the following complex numbers on an Argand diagram
a) \(z_1=4i\)
b) \(z_2=-2-2i\)
c) \(z_3=5-3i\)
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 \(z=rcisθ\)
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) \(z_1z_2\)
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)
\(z_1z_2 = 2 \sqrt2 \ cis \frac{π}{4} \times 2 \ cis \frac{π}{6} = 4 \sqrt2 \ cis \ \frac{5π}{12}\)
\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*}
Simplify the following expressions in Mod-Arg form
a) \(4cis \left( \frac{5π}{12} \right) \times 3cis \left( \frac{π}{4} \right)\)
b) \( \frac{3cis \left( \frac{π}{2} \right)}{4cis \left( – \frac{π}{6} \right)}\)
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
Simplify the following expressions in Mod-Arg form
a) \((4cis \left( \frac{π}{3} \right))^3\)
b) \((2cis \left( – \frac{2π}{5} \right))^3\)
a) \(64cis(π)=64(-1+0i)=-64\)
b) \(8cis \left(- \frac{6π}{5} \right) = 8cis \left( \frac{4π}{5} \right)\)
\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)
Express the following in exponential form
a) \(\sqrt3 -i\)
b) \(-2-2i\)
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*}
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) \((2e^{\frac{iπ}{6}})^{15}\)
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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