Complex Numbers Part 2

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Cartesian forms

• not covered, as it is from year 11 notes

Factorisation of Polynomials

• A polynomial function is a sum of multiples of powers of a variable
• E.g. $P(x)=5x^7-4x^5+3x^4-x^3+x+8$
• Degree/order of a polynomial is used to refer to the highest power of any variable in the polynomial
• The degree/order of the polynomial above is 7, because 7 is the highest power
• Usually, you write polynomials in order of decreasing power
  • This is not set in stone, there are multiple ways to do it

What this topic is about

• We already know the following about quadratic polynomials
• If a quadratic equation $a{x}^2+bx+c=0$, has solutions $x=p$ and $x=q$, the LHS admits a factorisation - $a(x-p)(x-q)=0$
  • i.e. solutions correspond to linear factors of the polynomial
• This chapter extends this idea to higher degree polynomials

Arithmetic with polynomials

• Polynomials can be added or subtracted by combining like terms and simplifying
• Polynomials can be multiplied using the distributive law
  • e.g. $(a+b)(c+d)=a(c+d)+b(c+d)$
• Polynomials can be divided using polynomial division

Division with integers

• Consider a positive integer $n$, e.g. 100, and a strictly smaller positive integer $p$, e.g. 23
• 100 is not divisible by 23, so there is a remainder $r$ when 100 is divided by 23
• Thus, $100=q\times 23+r$
  • $q$ must be the quotient
  • $r$ must always be smaller than the number
• When a polynomial $P(x)$ is divided by a polynomial $D(x)$, the degree of the remainder $R(x)$ will be strictly smaller than the degree of $D(x)$

Important Fact!

When a polynomial $P(x)$ is divided by a polynomial $D(x)$, the degree of the remainder $R(x)$ will be strictly smaller than the degree of $D(x)$

Thus, $P(x)=Q(x)D(x)+R(x)$, where $Q(x)$ is the quotient

Thus, $\frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$

• Compare with numerical division: when 100 is divided by 23, the remainder is 8, which is smaller than 23, and $100=4\times 23+8$
• Thus, $\frac{100}{23}=4+\frac{8}{23}$
• True algebraic fraction
  • Numerator has a lower degree than denominator
• Improper algebraic fraction
  • Numerator has a higher degree than denominator
• In certain later topics (i.e. using partial fractions for integration), it will be necessary to rewrite improper algebraic fractions in terms of proper ones
• If $P(x)$ and $(x-a)$ are polynomials then $P(x)=Q(x)(x-a)+r$
  • I.e. for any polynomials $P(x)$ and $(x-a)$, either $(x-a)$ is a factor of $P(x)$ or there is a polynomial multiple of $(x-a)$ that differs from $P(x)$ by a constant

The Remainder Theorem and the Factor Theorem

• Remember, if $P(x)$ and $(x-a)$ are polynomials then $P(x)=Q(x)(x-a)+r$

Remainder theorem:

for a polynomial $P(x)$ and a number $a$, the remainder when $P(x)$ is divided by $(x-a)$ is $P(a)$

• Proof
  • Let $r$ be the remainder when $P(x)$ is divided by $x-a$
  • Thus, for some polynomial $Q(x),P(x)=Q(x)(x-a)+r$
  • Hence, $P(x)=Q(a)(a-a)+r$, as required

Factor Theorem

For a polynomial $P(x)$, $(x-a)$ is a factor of $P(x)$ if and only if $P(a)=0$

• Proof 1
  • Suppose $(x-a)$ is a factor of $P(x)$
  • Then the remainder when $P(x)$ is divided by $(x-a)$ is 0
  • Thus, by the remainder theorem, $P(a)=0$
• Proof 2
  • Conversely, suppose that $P(a)=0$
  • Then the remainder when $P(x)$ is divided by $(x-a)$ is 0
  • Hence, $(x-a)$ is a factor of $P(x)$

Other stuff which is really useful to know (but not required in course content)

Theorem (the Fundamental Theorem of Algebra)

• Every real (real coefficients) polynomial equation of degree $n$ has exactly $n$ solutions (some of which may be repeated or complex)
• Every real polynomial $P(x)$ of degree $n\ge 1$ can be factorised as a product of $n$ linear factors $a(x-a_1)(x-a_2)\dots (x-a_n)$ where the $a_i$ are the zeroes of $P(x)$

Theorem (IMPORTANT!)

• If $z$ is a complex solution to a polynomial equation (with real coefficients), then so is $\overline{z}$
• Thus, complex solutions always come in conjugate pairs, so the total number of complex solutions (with non-zero imaginary part) is always even
• It also means that the total number of linear factors involving complex numbers is even
• Thus, any real polynomial of odd degree with have at least 1 real solution
• Odd degree: starts up/down and finishes down/up
• Even degree: starts up/down and finishes up/down

Theorem (Multiplicity of factors)

• Suppose that $(x-a{)}^n$ is a factor of $P(x)$. Then
  • If $n$ is odd, the graph of $y=P(x)$ cross the x-axis at $a$
  • If $n$ is even, the graph of $y=P(x)$ ‘touches’ but does not cross the x-axis at $a$

Polar Form of Complex numbers

• An exam question says $\text{Evaluate: }(2+3i)\star (1-7i)$

• The symbol $\star$ could be $+,-,÷,\times$

• Addition and subtraction in this form are much easier than multiplication and division

• Polar form is most natural form for multiplying/dividing complex numbers, or raising powers to complex numbers

• Multiplying by $i$ rotates the point 90˚ anticlockwise about the origin on the Argand plane

• When multiplying by a complex number $z=a+bi$, the things determining the corresponding geometric transformation are

  • the distance $\mid z\mid$ from the origin (the magnitude/modulus)
  • the angle $\theta$ from the positive real axis (the argument)

• How to rewrite $z$ so that $\mid z\mid$ and $\theta$ are explicit

  • $\cos \theta =\frac{a}{\mid z\mid }$
    • $a=\cos \theta \mid z\mid$
  • $\sin \theta =\frac{b}{\mid z\mid }$
    • $b=\sin \theta \mid z\mid$
  • $\theta =\tan \frac{b}{a}$

$z=\mid z\mid \cos \theta +i\mid z\mid \sin \theta =\mid z\mid (\cos \theta +i\sin \theta )=\mid z\mid \text{cis }\theta$

• where $\mid z\mid$ is the magnitude/modulus of $z$, i.e. the distance from the origin
• $\theta$ is the principal argument (angle between the vector for $z$ and the real axis) and $-\pi <\theta \le \pi$

Complex Conjugates in polar form

Recall for a complex number, $z=a+bi$

The complex conjugate is $\overline{z}=a-bi$

Then, if $z=r\text{ cis }\theta$, $\overline{z}=r\text{ cis }(-\theta )$

Multiplication of Complex Numbers

$r_1\text{ cis }{\theta }_1\times r_2\text{ cis }{\theta }_2=r_1{r}_2\text{ cis }({\theta }_1+{\theta }_2)$

Proof

LHS = r_{1} \text{ cis } \alpha \times r_{2 } \text{ cis } \beta \\ \\ = r_{1 } r_{2} [(\cos \beta + i \sin \beta)(\cos \alpha + i \sin \alpha )] \\ \\ = r_{1} r_{2}( \cos \alpha \cos \beta + i \cos \alpha \sin \beta + i \sin \alpha + \cos \beta + i^2 \sin \alpha \sin \beta) \\ \\ = r_{1} r_{2 } (\cos \alpha \cos \beta - \sin \alpha \sin \beta + i (\sin \alpha \cos \beta + \cos \alpha \sin \beta)) \\ \\ = r_{1} r_{2}[(\cos \alpha + \beta) + i \sin (\alpha + \beta)] \\ \\ = r_{1} r_{2} \text{ cis } (\alpha + \beta) \end{align}

Division of Complex Numbers

$r_1\text{ cis }{\theta }_1÷r_2\text{ cis }{\theta }_2=\frac{r_1}{r_2}\text{ cis }({\theta }_1-{\theta }_2)$

Regions in the Complex Plane

Set Notation

Analogue with real numbers: $\{x:1\le x<3,x\in \mathbb{R}\}$

i.e. all values of x such that those conditions are true

for complex numbers:

$z:|z|=4,z\in \mathbb{C}$

• circle centred at the origin, radius of 4
• some people call this the locus of $|z|=4$

$z:|z|\le 4,z\in \mathbb{C}$

• Just shade the interior of the circle to show that the numbers inside the circle are also included

$z:|z|<4,z\in \mathbb{C}$

• Show the circumference as a dotted line
• Shade the interior of the circle to show that the numbers inside the circle are also included

Represent $z:\text{arg }z=\frac{\pi }{6}$

• Make a line where the argument is $\frac{\pi }{6}$ radians
• Circle the origin, because $0+0i$ is excluded from the set because its argument is undefined

Sometimes the region represented by a set isn’t obvious $\{z:|z+8|=|z-4i|\}$

• One possible strategy: think of coordinate geometry, i.e. represent this image as x and y coordinates then rearrange units until it is in a suitable plot-able form
• Alternatively, observe that $|z-a|$ is the distance between the complex numbers $z$ and $a$

$n$th roots of 1

In the real world: $x^2=1,x=1,-1$ $x^3=1,x=1$ $x^4=1,x=1,-1$ $x^5=1,x=1$ $x^6=1,x=1,-1$ etc etc, alternating

However, $x^4=1$ has 4 solutions (1, -1, i, -i)

Consider $x^3=1$ which has just 1 real solution

We can solve using polynomial techniques, i.e. rewrite as $x^3-1=0$ and factorise LHS

$(x-1)(x^2+x+1)=0$

Using quadratic formula: $x=-\frac{1}{2}+\frac{\sqrt{3}}{2}i\text{ and }-\frac{1}{2}-\frac{\sqrt{3}}{2}i$

Theorem: including complex numbers, there are $n$ $nth$ solutions of 1. They have modulus 1 and are spaced at angular intervals of $\frac{2\pi }{n}$

More general theorem

Given a non-zero complex number $z$, the total number of $n$th roots of $z$ (including complex roots) is $n$

The roots al have the same modulus, and are spaced at angular intervals of $\frac{2\pi }{n}$, i.e. form the vertices of a regular n-gon

De Moirves Theorem

Recall that in polar form:

$r_1\text{ cis }{\theta }_1\times r_2\text{ cis }{\theta }_2=r_1{r}_2\text{ cis }({\theta }_1+{\theta }_2)$

Theorem: For any integer $n$ and complex number $r\text{ cis }\theta$

$(r\text{ cis }\theta {)}^n=r^n\text{ cis }(n\theta )$