Tag Archives: complex numbers

June 4, 2026 · 11:33 am

Complex Numbers – Roots of Unity

    \begin{equation*}z^n=1, z\in\mathbb{C}\end{equation}

In polar form 1=cis (2\pi)

Hence,

    \begin{equation*}z^n=cis(2\pi k), k\in\mathbb{Z}\end{equation}

    \begin{equation*}z^n=cis(\frac{2\pik}{n})\end{equation}

For example,

    \begin{equation*}z^4=1\end{equation}

    \begin{equation*}z^4=cis(2\pi k)\end{equation}

    \begin{equation*}z_k=cis(\frac{2\pi k}{4})=cis(\frac{\pi k}{2})\end{equation}

    \begin{equation*}z_0=1\end{equation}

    \begin{equation*}z_1=cis(\frac{\pi}{2})=i\end{equation}

    \begin{equation*}z_2=cis(\pi)=-1\end{equation}

    \begin{equation*}z_3=cis(\frac{3\pi}{2})=-i\end{equation}

The roots of unity are spread evenly (\frac{2\pi}{n} or \frac{360}{n} apart) around a circle of radius 1.

Properties of the roots of Unity

All the roots of unity can be generated by powers of a single root \omega=cis(\frac{2\pi}{n}). The roots form the sequence 1, \omega, \omega^2, \omega^3, ..., \omega^{n-1} Note: \omega \ne 1

For example,

    \begin{equation*}z^4=1\end{equation}

    \begin{equation*}\omega_1=cis(\frac{\pi}{2})\end{equation}

    \begin{equation*}\omega_1^2=cis(\pi)=-1=\omega_2\end{equation}

    \begin{equation*}\omega_1^3=cis(\frac{3\pi}{2})=-i=\omega_3\end{equation}

    \begin{equation*}\omega_1^4=cis(2\pi)=1=\omega_0\end{equation}

A root that can generate the remaining roots is a primitive root.

\omega^k is a primitive root if n and k are coprime.

The sum of the n^{th} roots of unity is always zero.

    \begin{equation*}\Sigma_{k=0}^{n-1}1+\omega+\omega^2+...+\omega^{n-1}=0\end{equation}

The product of the n^{th} roots of unity is

    \begin{equation*}\Pi_{k=0}^{n-1}\omega^k=(-1)^{n-1}\end{equation}

It is 1 when n is odd and -1 when n is even.

Example WATP 2024 Question 7a

(a) Evaluate (4\omega^2+3)(4\omega+3) where \omega is a complex root of unity, \omega\ne 1.

    \begin{equation*}(4\omega^2+3)(4\omega+3)=16\omega^3+12\omega^2+12\omega+9\end{equation}

    \begin{equation*}16\omega^3+12\omega^2+12\omega+9=16(1)+9(\omega^2+\omega+1)+3\omega^2+3\omega\end{equation}

    \begin{equation*}16\omega^3+12\omega^2+12\omega+9=16+9(0)+3(\omega^2+\omega)\end{equation}

    \begin{equation*}16\omega^3+12\omega^2+12\omega+9=16+3(-1)\end{equation}

    \begin{equation*}=13\end{equation}

Remember the sum of the roots is zero, \omega^2+\omega+1=0 hence \omega^2+\omega=-1

Leave a Comment

Filed under Algebra, Complex Numbers, Roots of Unity, Year 12 Specialist Mathematics

Tagged as , ,

February 11, 2026 · 3:28 am

Complex Numbers and Trig Idenities

My Year 12 Specialist Students are using complex numbers to prove trigonometric identities.

Things like

    \begin{equation*}sin(5\theta)=16sin^5(\theta) -20sin^3(\theta)+5sin(\theta) \end{equation}

Method 2 might be a little bit easier depending upon how your brain works.

Leave a Comment

Filed under Algebra, Binomial Expansion Theorem, Complex Numbers, Identities, Trig Identities, Trigonometry, Year 12 Specialist Mathematics

Tagged as , ,

October 1, 2024 · 7:31 am

Using De Moivre’s Theorem for Trigonometric Identities

We are going to use De Moivre’s theorem to prove trigonometric identities.

Remember, De Moivre’s Theorem

If z=r(cos(\theta)+isin(\theta)), then z^n=r^n(cos(n\theta)+isin(n\theta))

Or a shorter version z=rcis(\theta), then z^n=r^ncis(n\theta)

Now, let z=cos(\theta)+isin(\theta), find z+\frac{1}{z}

z+z^{-1}=cos(\theta)+isin(\theta)+cos(-\theta)+isin(-\theta)

Remember cos(\theta)=cos(\theta) and sin(-\theta)=-sin(\theta)

z+\frac{1}{z}=cos(\theta)+isin(\theta)+cos(\theta)-isin(\theta)

z+\frac{1}{z}=2cos(\theta)

It is the same for z^n+\frac{1}{z^n}

z^n+z^{-n}=cos(n\theta)+isin(n\theta)+cos(-n\theta)+isin(-n\theta)

z^n+\frac{1}{z^n}=2cos(n\theta)

Prove cos(2\theta)=2cos^2(\theta)-1
LHS=\frac{1}{2}(z^2+\frac{1}{z^2})
LHS=\frac{1}{2}(z^2+\frac{1}{z^2})+z\times\frac{1}{z}-z\times\frac{1}{z}
LHS=\frac{1}{2}(z^2+2z\times\frac{1}{z}+\frac{1}{z^2})-z\times\frac{1}{z}
LHS=\frac{1}{2}(z+\frac{1}{z})^2-1
LHS=\frac{1}{2}(2cos(\theta))^2-1
LHS=\frac{1}{2}(4cos^2(\theta))-1
LHS=2cos^2(\theta)-1
LHS=RHS

We can do something similar with sine.

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(-\theta)+isin(-\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(-\theta)+isin(-\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(\theta)-isin(\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-cos(\theta)+isin(\theta)

z-\frac{1}{z}=2isin(\theta)

Hence z^n-\frac{1}{z^n}=2isin(n\theta)

Prove sin(2\theta)=2sin(\theta)cos(\theta)
LHS=sin(2\theta)
LHS=\frac{1}{2i}(z^2-\frac{1}{z^2})
LHS=\frac{1}{2i}(z-\frac{1}{z})(z+\frac{1}{z})
LHS=\frac{1}{2i}(2isin(\theta)(2cos(\theta))
LHS=sin(\theta)2cos(\theta)
LHS=2sin(\theta)cos(\theta)
LHS=RHS

Let’s find an identity for cos(3\theta)

cos(3\theta)=\frac{1}{2}(z^3+\frac{1}{z^3})

=\frac{1}{2}(z^3+\frac{1}{z^3}+3z^2\times\frac{1}{z}+3z\times\frac{1}{z^2}-3z^2\times\frac{1}{z}-3z\times\frac{1}{z^2})

=\frac{1}{2}((z+\frac{1}{z})^3-3z-\frac{3}{z})

=\frac{1}{2}((z+\frac{1}{z})^3-3(z+\frac{1}{z}))

=\frac{1}{2}(2cos(\theta))^3-3(2cos(\theta)))

=\frac{1}{2}(8cos^3(\theta)-6cos(\theta))

=4cos^3(\theta)-3cos(\theta)

\therefore cos(3\theta)=4cos^3(\theta)-3cos(\theta)

And sin(3\theta)?

sin(3\theta)=\frac{1}{2i}(z^3-\frac{1}{z^3})

=\frac{1}{2i}(z^3-\frac{1}{z^3}-3z^2\times\frac{1}{z}+3z\times\frac{1}{z^2}+3z^2\times\frac{1}{z}-3z\times\frac{1}{z^2}

=\frac{1}{2i}((z-\frac{1}{z})^3+3z-\frac{3}{z})

=\frac{1}{2i}(2isin(\theta))^3+3(z-\frac{1}{z}))

=\frac{1}{2i}(-8isin^3(\theta)+6isin(\theta))

=-4sin^3(\theta)+3sin(\theta)

\therefore sin(3\theta)=3sin(\theta)-4sin^3(\theta)

Leave a Comment

Filed under Complex Numbers, Identities, Trig Identities, Trigonometry

Tagged as , ,

April 17, 2024 · 9:32 am

Sketching Subsets of the Complex Plane – Problem 2

I found this question on the madasmaths site – his resources are fabulous.

I am not sure I would have been able to do this in an exam.

I have split my solution into 6 images and there is a pdf version at the bottom

PDF version of my solution

Leave a Comment

Filed under Complex Numbers, Sketching Complex Regions

Tagged as , ,