# Help with the proof that the sum of all the roots of a complex number is zero

If a complex number $z \neq 0$ has n roots, then each root can be expressed as:

$$z_j=(\sqrt[n]{r}) e^{ {i (\theta +2\pi j) }/{n} }$$

For $j=0,1,2,...,n-1$

Thus, the summation of all the roots is

$$(\sqrt[n]{r})\left(e^{ i \theta /n }+e^{i (\theta+2\pi )/n} +e^{i (\theta+4\pi )/n}+...+e^{i [\theta+2\pi (n-1)]/n}\right)=(\sqrt[n]{r})e^{ \left( i \theta /n \right) }\left(1+e^{ \left( \frac{2\pi i }{n} \right) }+e^{ \left( \frac{4\pi i }{n} \right) }+...+e^{ \left( \frac{2\pi(n-1) i }{n} \right) }\right)=(\sqrt[n]{r})e^{ (i \theta /n )}\sum_{k=1}^{n}\left[e^{ \left( \frac{2\pi(k-1) i }{n} \right) }\right]$$

For the result to be zero, $$\sum_{k=1}^{n}\left[e^{ \left( \frac{2\pi(k-1) i }{n} \right) }\right]=0$$

Using the formula for the summation of a geometric progression, we have \begin{align} \dfrac{1-e^{2\pi i}}{1-e^\frac{2\pi i}{n}}=0 \end{align}

Finally, using that $i=e^{i\frac{\pi}{2}}$ we have

$$\dfrac{1-i^4}{1-i^{4/n}}=0$$

Given that the numerator is clearly zero, I just have to be sure that de denominator is different from zero.

How can I evaluate $(1-i^{4/n})$ and be sure that it is different from zero? Are all the steps correct?

If the complex number $e^{\left( \frac{2\pi i}{n} \right)}$ were transformed to polar form, we would have for the denominator:

$$1-e^\dfrac{2\pi i}{n}=1-\cos{\dfrac{2\pi}{n}}-i\sin{\dfrac{2\pi}{n}}$$

How can I evaluate this sine and cosine?

• There are various typos in the long displayed summation of all roots. There is also a typo in the next displayed line, $i$ is missing in the exponent. But the next displayed line recovers and inserts the $i$. – André Nicolas May 30 '15 at 3:44
• Thanks! I fixed these typos. – Yhoa May 30 '15 at 14:01

To show that $e^{2\pi i/n}\ne 1$ if $n\gt 1$, it is enough to observe that $\cos(\phi)=1$ only when $\phi$ is a multiple of $2\pi$.
Remark: As has been pointed out in an unfortunately currently deleted answer by lhf, the sum of the roots of a monic polynomial $x^n+ax^{n-1}+\cdots$ is $-a$. So if $n\gt 1$ then the sum of the roots of $x^n-z$ is $0$. No computation needed.