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?

• The square roots of a complex number will themselves be complex:

  • i.e. if $z^2=a+b\mathrm{i}$ then $z=c+d\mathrm{i}$

• We can then square ($c+d\mathrm{i}$) and equate it to the original complex number ($a+b\mathrm{i}$), as they both describe $z^2$:

  • $a+b\mathrm{i}={\left(c+d\mathrm{i}\right)}^2$

• Then expand and simplify:

  • $a+b\mathrm{i}=c^2+2cd\mathrm{i}+d^2{\mathrm{i}}^2$

  • $a+b\mathrm{i}=c^2+2cd\mathrm{i}-d^2$

• 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: $b=2cd$  (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 $\frac{b}{2d}=c$ 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 ($c+d\mathrm{i}$)

• Note that one root will be the negative of the other root

  • g. $c+d\mathrm{i}$ and  $-c-d\mathrm{i}$

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 $z = r(\cos \theta +\mathrm{isin}\theta )$then $\sqrt[n]{z} = {\left[r\left(\cos (\theta +2\pi k)+\mathrm{isin}(\theta +2\pi k)\right)\right]}^{\frac{1}{n}}$

  • This can be rewritten as $\sqrt[n]{z} = r^{\frac{1}{n}}\left(\cos \left(\frac{\theta + 2\pi k}{n}\right)+\mathrm{isin}\left(\frac{\theta + 2\pi k}{n}\right)\right)$

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

  • For $z^n=r{\mathrm{e}}^{\mathrm{i}\theta }$,  $z= \sqrt[n]{r}{\mathrm{e}}^{\frac{\theta +2\pi k}{n}\mathrm{i}}$

• 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

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

  •  $z={\mathrm{e}}^{\frac{2\mathrm{\pi }k}{n}\mathrm{i}}$where k = 0, 1, 2, …, n-1

    • This is given in the formula booklet

• These can be written 1, ω, ω², …, ω^n-1

  • Where $\omega ={\mathrm{e}}^{\frac{2\mathrm{\pi }}{n}\mathrm{i}}$

• The sum of the roots of unity is zero

  • $1+\omega +{\omega }^2+...+{\omega }^{n-1}=0$

• They can be used to find all the roots of the equation $z^n=r{\mathrm{e}}^{\mathrm{i}\theta }$

  • Find one root normally $\alpha =\sqrt[n]{r}{\mathrm{e}}^{\frac{\mathrm{i}\theta }{n}}$

  • Then the n distinct roots can be found by multiplying α by each root of unity

    • α, αω, αω², …, αω^n-1 

What are the geometric properties of roots of complex numbers?

• The n roots of any non-zero complex number $r{\mathrm{e}}^{\mathrm{i}\theta }$ 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

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)ω^n-1

  • If you write the vertex using exponential form $r{\mathrm{e}}^{\mathrm{i}\theta }$ it can make the multiplications easier

  • Then you can just add $\frac{2\mathrm{\pi }}{n}$ to the argument to get the next vertex

• Write all vertices in Cartesian form to get the coordinates