Complex numbers and ODEs: problem set 1 (on de Moivre's theorem and the roots of unity)

We write in terms of $x$ and $y$, giving \[ \sqrt{(x + 1)^2 + y^2} = 8 - \sqrt{(x - 1)^2 + y^2}. \]

Square both sides to get \[ (x + 1)^2 + y^2 = 64 - 16\sqrt{(x - 1)^2 + y^2} + (x-1)^2 + y^2. \]

Now simplify and move everything except the square root to one side. Square the equation equation and simplify to get \[ \frac{x^2}{16} + \frac{y^2}{15} = 1, \]

which is an equation of an ellipse centred at the origin.

We first write \[ \frac{z-4}{z-1} = \frac{x^2 + y^2 - 4x + 5}{(x-1)^2 + y^2} + i \frac{y}{(x-1)^2 + y^2}, \]

by applying the trick of multiplying by the complex conjugate of the denominator. The argument of the above expression must be $3\pi/2$, so that means that the real part must be zero and the imaginary part must be negative. The first requirement sets \[ \left(x - \frac{5}{2}\right)^2 + y^2 = \frac{9}{4}, \]

after simplifying. Thus, the graph is a semi-circle in the lower half-plane of radius $3/2$ centred at the point $(5/2, 0)$.

The proof of the first portion is straightforward. Once we have the expression, then we wish to solve for $\cos\theta$. Application of the quadratic formula gives \[ \cos\theta = \frac{1}{2} \left[ 1 \pm \sqrt{1 - \frac{1}{2}(1-\cos 4\theta)}\right]. \]

Note that the discriminant is always positive. Which branch of the square root you select can be argued from intuition. Notice that if we put $\theta = \pi/8$ and $\theta = 3\pi/8$ into the above, then the right hand sides are identical except for a change of sign. However, because $\cos(\pi/8) > \cos(3/\pi/8)$, then the positive sign must be associated with $\theta = \pi/8$ and the negative sign to $\theta = 3\pi/8$. With this in mind, we get after simplifying, \[ \cos \pi/8 = \sqrt{\frac{2 + \sqrt{2}}{4}} \qquad \cos 3\pi/8 = \sqrt{\frac{2 - \sqrt{2}}{4}} \]

The $n$ roots are given by \begin{equation} z = a^{1/n} e^{i(\theta + 2\pi k)/n}, \end{equation}

for $k = 1, 2, \ldots, n - 1$. We now wish to show that \begin{equation} \sum_{k=0}^{n-1} \sin \left( \frac{2 \pi k}{n} \right) = 0, \end{equation}

and similarly for the cosine terms. Let us take the case of $n$ being even first. Then the above sum is \[ \sin(0) + \sin(\pi) + \sum_{k=1}^{n/2} \sin \left( \frac{2 \pi k}{n} \right) + \sum_{k=-(n/2)}^{-1} \sin \left( \frac{2 \pi k}{n} \right). \]

We can use the oddness property of sines to then write \[ \sin(0) + \sin(\pi) + \sum_{k=1}^{n/2} \sin \left( \frac{2 \pi k}{n} \right) + \sum_{k=1}^{n/2} \sin \left(-\frac{2 \pi k}{n} \right) = 0. \]

The proof of the odd version is done similarly and the proof for the cosine terms is done by profiting from the oddness of cosines about the $y$ axis.

Solving the complex equation, we get \[ z = \exp i\left(\frac{2k+1}{2n+1}\right), \]

for $k = -n, -n+1, \ldots, n-1, n$. Applying the result of the last part of the question, we conclude that the sum of the roots must be zero. The real part of the sum then implies the desired result.

Splitting the terms and taking the $n$th root gives \[ (z - 1) = A(z + 1), \]

where \[ A = \exp i\left(\frac{\pi + 2\pi k}{n}\right), \]

for $k = 0, 1, 2, \ldots, n -1$. Solving for $z$ now gives \[ z = \frac{1+A}{1-A} = i\cot\left( \frac{\pi + 2\pi k}{2n} \right). \]

\[ S = \sum_{r=1}^n \binom{n}{r} \sin (2r\theta) = \Im \left[ \sum_{r=0}^n \binom{n}{r} e^{2ir\theta} - 1 \right] \]

By the Binomial Theorem, \[ (1 + e^{2i\theta})^n = \sum_{r=0}^n \binom{n}{r} e^{2ir\theta}. \]

Thus we have (removing the $-1$ since it does not factor into the imaginary part) \begin{align*} S &= \Im \left[ (\left(1 + e^{2i\theta}\right)^n \right] \\ &= \Im \left[\left( (\cos 2\theta + 1) + i\sin 2\theta\right)^n\right] \\ &= \Im \left[\left( (2\cos^2 \theta -1 + 1) + 2i\sin \theta\cos\theta\right)^n\right]\\ &= \Im \left[2^n \cos^n\theta \left( \cos \theta + i\sin \theta\right)^n\right]\\ &= \Im \left[2^n \cos^n\theta \left( \cos n\theta + i\sin n\theta\right)\right]\\ &= 2^n \cos^n\theta \sin \theta \end{align*}

where we have used the result that $\cos 2\theta = 2\cos^2 \theta - 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$.