How can one prove that the product of all the roots of a complex equation is the same as one root to the power of equation?
Example: $z^n=a+bi$ has $n$ roots (from de Moivre's formula), prove that their product is same as any single root to the power of $n$.
\begin{align*}
z^n&=a+bi\\
z(k)^n&=[z(1)z(2)…z(n)]\\
&=a+bi\\
n\in \mathbb{Z}^+&,
\quad k\in n
\end{align*}
it is just an assumption made on special case
Assumption was made on equation $z^3=-\sqrt{3}+i$ where roots are:
$\cos(70)+j\sin(70)$, $\cos(250)+j\sin(250)$, $\cos(370)+j\sin(370)$, so the product of all is $-\sqrt{3}+i$. However, if I take any single to the power of 2 and multiply it by one of the rest, the result is different.
Best Answer
By multiplying out $P(z) = (z-a_1)(z-a_2)\cdots(z-a_n)$, we see that the product of the solutions $a_1,a_2,\ldots,a_n$ to $P(z)=0$ is $(-1)^n$ times the constant term of $P$.
In particular, if $n$ is odd, then the product of the solutions to $z^n-(a+bi)$ is $a+bi$.
If $n$ is even, then your claim is not true, due to a minus sign.