Product of all complex roots of $z^n=a+bi$

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.

• I'm not so sure of that. Consider $z^{2}=-1$ then the 2 root are $i,-i$. Their product is $1$ but $i^{2}=-1$. – Gina Jan 7 '14 at 7:58
• It's not true. There are complex equations other than $z^n = a+bi$. What is true is that the product of the roots of a polynomial of degree $n$ is $(-1)^n$ times the constant term of that polynomial. – Robert Israel Jan 7 '14 at 7:58
• @RobertIsrael, added the reason of wrong assumption for genial case. – user119510 Jan 7 '14 at 8:23

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.
• can you explain in details until $(-1)^n$ – user119510 Jan 7 '14 at 10:50
• @MotherLand Try it out with small degrees first. One way to look at it is to take $P(z)$ modulo $z$. Then we get $(-a_1)(-a_2)\cdots(-a_n) = (-1)^n a_1 a_2\cdots a_n$ for the constant term. – Slade Jan 7 '14 at 20:39