Root of complex polynomial

JML · #1 $Old$ October 28th, 2009, 10:37 PM

I was asked if it is possible to solve this question with the complex exponential:

$(z + i)^5 - (z-i)^5=0$

I can see how this is solved by expanding out, but is there an easier or more systematic way to go about these problems?

Bruno J. · #2 $Old$ October 28th, 2009, 10:59 PM

Well it is

$(z+i)^5=(z-i)^5$

Taking fifth roots on both sides, we have

$\omega(z+i)=z-i$

where $\omega^5=1$ is any fifth root of unity. Thus

$z(1-\omega)=i(1+\omega)$

or $z=\frac{i(1+\omega)}{1-\omega}$ .