Roots of Complex Numbers

How do I find the square root of a complex number?

The square roots of a complex number will themselves be complex:
- i.e. if $z^{2} = a + b i$ then $z = c + d i$
We can then square ( $c + d i$ ) and equate it to the original complex number ( $a + b i$ ), as they both describe $z^{2}$ :
- $a + b i = {(c + d i)}^{2}$
Then expand and simplify:
- $a + b i = c^{2} + 2 c d i + d^{2} i^{2}$
- $a + b i = c^{2} + 2 c d i - d^{2}$
As both sides are equal we are able to equate real and imaginary parts:
- Equating the real components: $a = c^{2} - d^{2}$ (1)
- Equating the imaginary components: $b = 2 c d$ (2)
These equations can then be solved simultaneously to find the real and imaginary components of the square root
- In general, we can rearrange (2) to make $\frac{b}{2 d} = c$ and then substitute into (1)
- This will lead to a quartic equation in terms of d; which can be solved by making a substitution to turn it into a quadratic
The values of $d$ can then be used to find the corresponding values of $c$ , so we now have both components of both square roots ( $c + d i$ )
Note that one root will be the negative of the other root
- g. $c + d i$ and $- c - d i$

How do I use de Moivre’s Theorem to find roots of a complex number?

De Moivre’s Theorem states that a complex number in modulus-argument form can be raised to the power of n by

Raising the modulus to the power of n and multiplying the argument by n

When in modulus-argument (polar) form de Moivre’s Theorem can then be used to find the roots of a complex number by
- $k = 0, 1, 2, \dots, n - 1$
- Recall that adding 2π to the argument of a complex number does not change the complex number
- Therefore we must consider how different arguments will give the same result

Taking the nth root of the modulus and dividing the argument by n
If $z = r (\cos θ + isin θ)$ then $\sqrt[n]{z} = {[r (\cos (θ + 2 π k) + isin (θ + 2 π k))]}^{\frac{1}{n}}$
This can be rewritten as $\sqrt[n]{z} = r^{\frac{1}{n}} (\cos (\frac{θ + 2 π k}{n}) + isin (\frac{θ + 2 π k}{n}))$

This can be written in exponential (Euler’s) form as

For $z^{n} = r e^{i θ}$ , $z = \sqrt[n]{r} e^{\frac{θ + 2 π k}{n} i}$

The nth root of complex number will have n roots with the properties:
- The five roots of a complex number raised to the power 5 will create a regular pentagon on an Argand diagram
- The eight roots of a complex number raised to the power 8 will create a regular octagon on an Argand diagram
- The n roots of a complex number raised to the power n will create a regular n-sided polygon on an Argand diagram

The modulus is $\sqrt[n]{r}$ for all roots
There will be n different arguments spaced at equal intervals on a circle centred about the origin
This creates some geometrically beautiful results

Examiner Tip

de Moivre's theorem makes finding roots of complex numbers very easy, but you must be confident converting from Cartesian form into Polar and Euler's form first
- You can use your calculator to convert between forms

Worked example

Find the square roots of 5 + 12i, giving your answers in the form a + bi.

pKIYVs8H_al-fm-1-2-4-roots-of-cn-we-solu-1

Solve the equation

z^{3} = - 4 + 4 \sqrt{3} i

giving your answers in the form r (cosθ + isinθ).

LAXY4Oza_al-fm-1-2-4-roots-of-cn-we-solu-2

Roots on an Argand Diagram

What are roots of unity?

Roots of unity are solutions to the equation $z^{n} = 1$ where n is a positive integer
For the equation $z^{n} = 1$ there are n roots of unity
- $z = e^{\frac{2 π k}{n} i}$ where k = 0, 1, 2, …, n-1
  - This is given in the formula booklet
These can be written 1, ω, ω², …, ωⁿ^-1
- Where $ω = e^{\frac{2 π}{n} i}$
The sum of the roots of unity is zero
- $1 + ω + ω^{2} + . . . + ω^{n - 1} = 0$
They can be used to find all the roots of the equation $z^{n} = r e^{i θ}$
- Find one root normally $α = \sqrt[n]{r} e^{\frac{i θ}{n}}$
- Then the n distinct roots can be found by multiplying α by each root of unity
  - α, αω, αω², …, αωⁿ^-1

What are the geometric properties of roots of complex numbers?

The n roots of any non-zero complex number $r e^{i θ}$ lie on a circle on an Argand diagram

The centre will be the origin
The radius will be $\sqrt[n]{r}$

The n roots of unity lie on the unit circle centred about the origin
Regular polygons can be created by joining consecutive roots of a complex number with straight lines

edexcel-al-fm-cp-1-2-4-unity

How can I use roots of unity to solve geometric problems?

Roots of unity can be used to solve problems involving regular polygons centred about the origin
Coordinates of vertices (x, y) can be considered as complex numbers x + yi
If you know one vertex (x, y) you can find the others by multiplying the complex number representing the given vertex by each root of unity

x + yi, (x + yi)ω, (x + yi)ω², …, (x + yi)ωⁿ^-1
If you write the vertex using exponential form $r e^{i θ}$ it can make the multiplications easier
Then you can just add $\frac{2 π}{n}$ to the argument to get the next vertex

Write all vertices in Cartesian form to get the coordinates

Examiner Tip

You can use your calculator to convert between polar and cartesian forms which may speed up your working
- Just be aware of questions that may ask you to not use “calculator technology” where you need to show full working (but can still use calculator to check!)

Worked example

An equilateral triangle has its centre at the origin of a Cartesian plane. One of its vertices is at the point $(9, - 3)$ . Find the coordinates of the other two vertices.

al-fm-1-2-4-roots-on-argand-diagram-we-s

Roots of Complex Numbers (Edexcel A Level Further Maths: Core Pure)

Revision Note

Roots of Complex Numbers

How do I find the square root of a complex number?

How do I use de Moivre’s Theorem to find roots of a complex number?

Examiner Tip

Worked example

Roots on an Argand Diagram

What are roots of unity?

What are the geometric properties of roots of complex numbers?

How can I use roots of unity to solve geometric problems?

Examiner Tip

Worked example

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Author: Amber