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

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

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

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- Oct 28th 2009, 08:37 PMJMLRoot of complex polynomial
I was asked if it is possible to solve this question with the complex exponential:

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

I can see how this is solved by expanding out, but is there an easier or more systematic way to go about these problems? - Oct 28th 2009, 08:59 PMBruno J.
Well it is

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

Taking fifth roots on both sides, we have

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

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

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

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