Category Archives: Complex Numbers

January 22, 2026 · 9:27 am

Complex Loci Question

A sketch of the locus of a complex number z is shown above, determine the maximum value of arg(z) correct to two decimal places where 0\le z \le 2\pi

Draw tangent lines from the origin to the circle.

Remember tangent lines are perpendicular to the radii

The maximum argument is this angle

I am going to find the angle in two sections

From the diagram the radius of the circle is 2 and the centre is (4, 3). Hence the distance from the origin to the centre is 5.

    \begin{equation*}sin(\theta_1)=\frac{2}{5}\end{equation}

    \begin{equation*}\theta_1=0.412\end{equation}

    \begin{equation*}sin(\theta_2)=\frac{3}{5}\end{equation}

    \begin{equation*}\theta_2=0.644\end{equation}

Hence maximum arg(z)=1.06

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October 18, 2024 · 7:06 am

Fibonacci Sequence – Finding the Closed Form

I have been reading An Imaginary Tale – The Story of \sqrt{-1} by Paul J Nahin, which is fabulous. There was a bit in chapter 4 where he found the closed form of the generalised Fibonacci sequence. I thought it would be a good exercise to find the closed from of the Fibonacci sequence.

Just to remind you, the Fibonacci sequence is

1, 1, 2, 3, 5, 8, 13, 21, ...

and it is defined recursively

    \begin{equation*}T_{n+2}=T_{n+1}+T_n, T_0=1, T_1=1\end{equation}

That is, the next term is the sum of the two previous terms, i.e.

    \begin{equation*}T_3=T_2+T_1=1+1=2\end{equation}

Now the starting off point is slightly dodgy as it involves and educated guess as Paul Nahin writes,

How do I know that works? Because I have seen it before, that’s how! […] There is nothing dishonourable about guessing correct solutions – indeed, great mathematicians and scientists, are invariable great guessers – just as long as eventually the guess is verified to work. The next time you encounter a recurrence formula, you can guess the answer too because then you will have already seen how it works.

We start with T_n=kz^n

This means T_{n+2}=T_{n+1}+T_n is kz^{n+2}=kz^{n+1}+kz^n

    \begin{equation*}kz^{n+2}=kz^{n+1}+kz^n\end{equation}

    \begin{equation*}kz^n(z^2-z-1)=0\end{equation}

    \begin{equation*}z^2-z-1=0\end{equation}

    \begin{equation*}z=\frac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2}\end{equation}

\therefore z=\frac{1+\sqrt{5}}{2} or z=\frac{1-\sqrt{5}}{2}

Hence T_n=k_1(\frac{1+\sqrt{5}}{2})^n+k_2(\frac{1-\sqrt{5}}{2})^n and we can use the initial conditions T_0=1 and T_1=1 to find k_1 and k_2

When n=0, T_0=1

(1)   \begin{equation*}1=k_1+k_2\end{equation*}

When n=1, T_1=1

(2)   \begin{equation*}1=k_1(\frac{1+\sqrt{5}}{2})+k_2(\frac{1-\sqrt{5}}{2})\end{equation*}

From equation 1, k_2=(1-k_1), substitute into equation 2

    \begin{equation*}1=k_1(\frac{1+\sqrt{5}}{2})+(1-k_1)(\frac{1-\sqrt{5}}{2})\end{equation}

    \begin{equation*}1=k_1(\frac{1+\sqrt{5}}{2})+\frac{1-\sqrt{5}}{2}-k_1(\frac{1-\sqrt{5}}{2})\end{equation}

    \begin{equation*}1=k_1\sqrt{5}+\frac{1-\sqrt{5}}{2}\end{equation}

    \begin{equation*}1-(\frac{1-\sqrt{5}}{2})=\sqrt{5}k_1\end{equation}

    \begin{equation*}k_1=\frac{1}{\sqrt{5}}(\frac{1}{2}+\frac{\sqrt{5}}{2})\end{equation}

    \begin{equation*}k_1=\frac{1}{2}(\frac{1}{\sqrt{5}}+1)\end{equation}

    \begin{equation*}k_1=\frac{1+\sqrt{5}}{2\sqrt{5}}\end{equation}

    \begin{equation*}k_2=1-\frac{1+\sqrt{5}}{2\sqrt{5}}\end{equation}

    \begin{equation*}k_2=-(\frac{1-\sqrt{5}}{2\sqrt{5}})\end{equation}

\therefore T_n=(\frac{1+\sqrt{5}}{2\sqrt{5}})(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2\sqrt{5}})(\frac{1-\sqrt{5}}{2})^n

T_n=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1})

Does it work?

Remember the sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...

If n=5, T_n=8

    \begin{equation*}T_5=\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^6-\frac{1-\sqrt{5}}{2})^6)\end{equation}

As you can see it works!

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Filed under Complex Numbers, Fibonacci, Fibonacci Sequence, Interesting Mathematics, Sequences

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)

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August 23, 2024 · 7:41 am

Sketching in the Complex Plane Using a TI-nspire CX 11

Let R be the region of the complex plane where the inequalities |z-i|\le2 and |z-\bar{z}|\ge3 hold simultaneously.

First find the Cartesian equations.

Finding the symbols


The conj(z) is found under
Menu – Number – Complex Number Tools – Complex Conjugate

Second, sketch each of the functions.

The section that is shaded twice is our region.

Determine the minimum value of Re(z) in R.

We can find the point of intersection between the circle and the line.

Re(z)=-1.94

Or if you want exact values

Use the Solve Systems of Equations tool – Menu – Algebra – Solve Systems of Equations – Solve Systems of Equations.

Re(z)=-\frac{\sqrt{15}}{2}

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Filed under Complex Numbers, Sketching Complex Regions, TI nspire CX 11

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

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