Show that if $z = e^{i\theta}$ is a solution to $0 = z^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0$, ...

Question

Show that if $z = e^{i\theta}$ is a solution to $0 = z^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0$ [1] where all $a_i$ are real, then $0 = a_{n-1}\sin\theta + a_{n-2}\sin2\theta + \cdots + a_0 \sin n\theta$ [2]

So far I have realized that [2] states that the complex polynomial $a_0z^n +a_1z^{n-1}+\cdots a_{n-1}z$ [3] is real. I am trying to convert [1] to [3] but have not had any luck so far. I've tried some other stuff as well; can anyone solve the problem?

Thank you so much everyone for your solutions! I woke up this morning and in a flash of inspiration figured out the solution. Looks like my proof agrees with yours'.

Answer 1

Note that since $$ 0= e^{in\theta}+ a_{n-1}e^{i(n-1)\theta}+\cdots + a_0 $$ Then, dividing by $e^{in\theta}$ we get $$ 0=1+a_{n-1}e^{-i\theta}+\cdots +a_0e^{-in\theta} $$ Since $e^{ix}=\cos(x)+i\sin(x)$ for every real $x$, then after grouping some terms we have $$ 0=(1+a_{n-1}\cos(-\theta)+\cdots +a_0\cos(-n\theta))+i(a_{n-1}\sin(-\theta)+\cdots +a_0\sin(-n\theta)) $$ Since all of the $a_k$ are real, this expression is in the form $a+ib$ where $a,b\in\mathbb{R}$ and since it is equal to zero, we should have $b=0$, that is, $$ 0=a_{n-1}\sin(-\theta)+\cdots +a_0\sin(-n\theta)) $$ And using that $\sin$ is an odd function, we have the result.

Answer 2

For notational convenience, set $a_n = 1$. Then the given equation may be written

$\sum_{m = 0}^n a_m z^m = 0; \tag{1}$

next, recall that $(\cos m \theta + i\sin m \theta) = e^{i m \theta} = (e^{i \theta})^m$ by de Moivre's formula. Inserting $z = e^{i \theta}$ into (1), it becomes

$\sum_{m = 0}^n a_m e^{i m \theta} =0, \tag{2}$

so that now if we multiply through by $e^{-i n \theta}$ we obtain

$\sum_{m = 0}^n a_m e^{i (m - n) \theta} = 0, \tag{3}$

or

$\sum_{m = 0}^n a_m (\cos (m - n) \theta + i\sin (m- n) \theta) = 0. \tag{4}$

Notice that the $m = n$ term in (4) is $1$; as for the rest, taking the imaginary part yields

$\sum_{m = 0}^{n - 1} a_m \sin (n - m) \theta = 0, \tag{5}$

where the fact that $\sin(- \alpha) = -\sin \alpha$ has been used in the transition 'twixt (4) and (5). It is easy to see that (5) is the requisite sum. QED.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!