Simpyfying the expression $(1 + \cos \alpha + i\sin \alpha)^k + (1 + \cos \alpha - i\sin \alpha)^k$

Question

I came across the following question the other day and was wondering if my solution is the correct solution. Here is the question.

Prove that

$$ (1 + \cos \alpha + i\sin \alpha)^k + (1 + \cos \alpha - i\sin \alpha)^k = 2^{k+1} \cos \left( \frac{k\alpha}{2} \right) \cos^k \left( \frac{\alpha}{2} \right) $$

My solution is:

$$ (1 + \cos \alpha + i\sin \alpha)^k + (1 + \cos \alpha - i\sin \alpha)^k = 2^{k+1} \cos \left( \frac{k\alpha}{2} \right) \left| \cos^k \frac{\alpha}{2} \right| $$

The absolute value is obtained if, say, $z = (1 + \cos \alpha) + i\sin \alpha $, then after some trig substitution one gets.

$$ |z| = \sqrt{ 4 \cos^2 \frac{\alpha}{2} } $$ $$ |z| = 2\left| \cos^k \frac{\alpha}{2} \right| $$

Finding Arg $z = \dfrac{\alpha}{2}$, one gets:

$$ z = 2 \left|\cos \frac{\alpha}{2} \right| \left( \cos \frac{\alpha}{2} + i \sin \frac{\alpha}{2} \right) $$ $$ \overline{z} =2 \left|\cos \frac{\alpha}{2} \right| \left( \cos \frac{\alpha}{2} - i \sin \frac{\alpha}{2} \right) $$

Using de Moivre's theorem and simplifying I get

$$ (z)^k + (\overline{z})^k = 2^{k+1} \cos \left( \frac{k\alpha}{2} \right) \left| \cos^k \frac{\alpha}{2} \right| $$

So is my solution correct using the absolute value or am I missing something?

Answer 1

Consider the expression $1 + \cos \alpha + i \sin \alpha$. We can represent this expression in polar form as $r (\cos \theta + i \sin \theta)$, where $r$ is the modulus and $\theta$ is the argument of the complex number. To find $r$, we calculate the magnitude of the complex number:

\begin{align*} r &= \sqrt{(1 + \cos \alpha)^2 + \sin^2 \alpha} \\ &= \sqrt{2 + 2 \cos \alpha} \\ &= \sqrt{2(1 + \cos \alpha)} \\ &= \sqrt{2(2 \cos^2 \left(\frac{\alpha}{2}\right))} \\ &= 2 \cos \left(\frac{\alpha}{2}\right). \end{align*}

The argument $\theta$ is determined by the tangent of the angle, which is given by:

\begin{align*} \tan \theta &= \frac{\sin \alpha}{1 + \cos \alpha} \\ &= \frac{2 \sin \left(\frac{\alpha}{2}\right) \cos \left(\frac{\alpha}{2}\right)}{2 \cos^2 \left(\frac{\alpha}{2}\right)} \\ &= \tan \left(\frac{\alpha}{2}\right), \end{align*}

implying that $\theta = \frac{\alpha}{2}$.

Similarly, for the expression $1 + \cos \alpha - i \sin \alpha$, we find it can also be represented in polar form as $r (\cos (-\theta) + i \sin (-\theta)) = r (\cos \theta - i \sin \theta)$, using the same value of $r$ and $\theta$.

Using De Moivre's theorem, we can express the sum of the $k$th powers of these complex numbers as:

\begin{align*} (1 + \cos \alpha + i \sin \alpha)^k + (1 + \cos \alpha - i \sin \alpha)^k &= r^k (\cos k\theta + i \sin k\theta + \cos k\theta - i \sin k\theta) \\ &= 2 r^k \cos k\theta \\ &= 2 \cdot 2^k \cos^k \left(\frac{\alpha}{2}\right) \cos \left(\frac{k\alpha}{2}\right) \\ &= 2^{k+1} \cos^k \left(\frac{\alpha}{2}\right) \cos \left(\frac{k\alpha}{2}\right). \end{align*}

Answer 2

There's a quick way to test if there's a problem. We can just substitute some values for $k$ and $\alpha$, particularly ones where the two final answers will differ, and see which produces the correct answer. In particular, if we let $k = 1$, and for the sake of nice trig values, but $\cos(\alpha/2) < 0$, take $\alpha = 7\pi/3$, then $$(1 + \cos \alpha + i\sin \alpha)^k + (1 + \cos \alpha - i\sin \alpha)^k = 1 + \frac{1}{2} + i \frac{\sqrt{3}}{2} + 1 + \frac{1}{2} - i\frac{\sqrt{3}}{2} = 3,$$ and $$2^{k+1} \cos \left( \frac{k\alpha}{2} \right) \cos^k \left( \frac{\alpha}{2} \right) = 4\left(-\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) = 3.$$ The original formula correctly evaluates this case, at least, while yours will differ by a sign.

The error here appears to be when you calculate the argument to be $\frac{\alpha}{2}$. You don't really include your reasoning here, so it's difficult to pick specifically where you went wrong. However, you can observe that your next two equations work fine if you remove the absolute values. In particular, \begin{align*} 2 \cos \left( \frac{\alpha}{2} \right) \left( \cos \frac{\alpha}{2} + i \sin \frac{\alpha}{2} \right) &= 2 \cos^2\left(\frac{\alpha}{2}\right) + 2i\cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\alpha}{2}\right) \\ &= 1 + \cos\left(2\frac{\alpha}{2}\right) + i\sin\left(2\frac{\alpha}{2}\right) \\ &= z, \end{align*} and the conjugate can be found by taking the conjugate of both sides. So, somewhere in that step, this erroneous absolute value crept in!

EDIT: Here's a correct argument from the polar form perspective, in response to several comments from Stephan. Stephan's argument is correct: the modulus of $z$ is $$|z| = 2\left|\cos \frac{\alpha}{2}\right|.$$ ADAM's calculation of the argument is (mostly) correct up to a point:

\begin{align*} \tan \theta &= \frac{\sin \alpha}{1 + \cos \alpha} \\ &= \frac{2 \sin \left(\frac{\alpha}{2}\right) \cos \left(\frac{\alpha}{2}\right)}{2 \cos^2 \left(\frac{\alpha}{2}\right)} \\ &= \tan \left(\frac{\alpha}{2}\right). \end{align*}

The only thing I would caution so far is that we need to account for division by zero. The first step works only if $\cos \alpha \neq -1$, i.e. when $\alpha \neq \pi + 2k\pi$ for all $k \in \Bbb{Z}$. The remaining steps are all OK, but we just need to keep this (very possible) case in mind once we finish the argument.

The next step, saying $\tan \theta = \tan \frac{\alpha}{2} \implies \theta = \frac{\alpha}{2}$ is false, even modulo $2\pi$, because $\tan$ has a period of $\pi$. It is entirely possible that $\theta = \pi + \frac{\alpha}{2}$ instead (or $-\pi + \frac{\alpha}{2}$, but this is the same modulo $2\pi$). Note that adding (or subtracting) $\pi$ from the argument simply multiplies the complex number by $-1$. So, choosing the argument to be $\frac{\alpha}{2}$ either works, or introduces an erroneous factor of $-1$.

How do we tell which is which? We require that $\cos \theta + i \sin \theta$ have real and positive parts whose signs are the same as $\cos \frac{\alpha}{2} + i \sin\frac{\alpha}{2}$. Note that $1 + \cos \alpha$ is always non-negative, and indeed only $0$ in our troubling $\pi + 2k\pi$ case. So, continuing to exclude this case, we require that our $\theta$ be such that $\cos(\theta) > 0$, even when $\cos \frac{\alpha}{2} < 0$. So, in the case where $\cos \frac{\alpha}{2} < 0$, the argument is not $\frac{\alpha}{2}$, but $\pi + \frac{\alpha}{2}$. In this case, $$\cos\theta + i\sin\theta = \cos\left(\pi + \frac{\alpha}{2}\right) + i\sin\left(\pi + \frac{\alpha}{2}\right) = -\cos\frac{\alpha}{2} - i\sin\frac{\alpha}{2}.$$ When $\cos \frac{\alpha}{2} = 0$, then $\alpha$ must be an odd multiple of $\pi$, which puts us in the (up until now) neglected $\cos \alpha = -1$ case. In this case, $\sin \alpha = 0$, and the modulus is $0$. In this case, the argument is undetermined, and any $\theta$ will work.

So, we see that: \begin{align*} z &= \begin{cases} 2\left|\cos \frac{\alpha}{2}\right|\left(\cos\frac{\alpha}{2} + i\sin\frac{\alpha}{2}\right) & \text{if } \cos\frac{\alpha}{2} > 0 \\ 2\left|\cos \frac{\alpha}{2}\right|\left(\cos\left( \pi + \frac{\alpha}{2} \right) + i\sin\left( \pi + \frac{\alpha}{2}\right)\right) & \text{if } \cos\frac{\alpha}{2} < 0 \\ 0 & \text{if } \cos \frac{\alpha}{2} = 0. \end{cases} \\ &= 2\left|\cos \frac{\alpha}{2}\right| \cdot \operatorname{sgn}\left(\cos \frac{\alpha}{2}\right) \cdot \left(\cos\frac{\alpha}{2} + i\sin\frac{\alpha}{2}\right) \\ &= 2\left(\cos \frac{\alpha}{2}\right) \left(\cos\frac{\alpha}{2} + i\sin\frac{\alpha}{2}\right), \end{align*} where $$\operatorname{sgn}(x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0.\end{cases}$$ In particular, the last step follows from the fact that $\operatorname{sgn}(x) \cdot |x| = x$ for all $x \in \Bbb{R}$.

Note that the final step is not always the polar form of $z$: the "modulus" out the front can be negative. However, this takes us back to the right path. As you can see, the polar form of $z$ is close, but not quite the best representation for it in order to solve the problem.