IDK if this is the best way...but we can first simplify it like this:
z49 + z50 + z51 + z52 + z53
Factor z49 out of the first three terms and z51 out of the last two terms
= z49(1 + z + z2) + z51(z + z2)
Since z2 + z + 1 = 0, 1 + z + z2 = 0 and z + z2 = -1
= z49( 0 ) + z51( -1 )
= -z51
Now by the quadratic formula, \(z\ =\ \frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}\ =\ \frac{-1\pm\sqrt{-3}}{2}\ =\ -\frac12\pm \frac{\sqrt3}{2}i\)
Let's pick \(z\ =\ -\frac12 + \frac{\sqrt3}{2}i\) (If we picked \(z\ =\ -\frac12 - \frac{\sqrt3}{2}i\) we would get the same answer)
Now let's re-express z to be in the form \(r(\cos\theta+i\sin\theta)\) so that we can use DeMoivre's Theorem.
By the Pythagorean Theorem,
\(r^2\ =\ (-\frac12)^2+(\frac{\sqrt3}{2})^2\ =\ 1\) so taking the positive sqrt, we get \(r \ =\ 1\)
An angle which has a cos of \(-\frac12\) and a sin of \(\frac{\sqrt{3}}{2}\) is \(\frac{2\pi}{3}\), so let \(\theta\ =\ \frac{2\pi}{3}\)
And so...
\(z\ =\ 1(\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3}))\) (we can check in a calculator that this does equal \( -\frac12 + \frac{\sqrt3}{2}i\) )
Then by DeMoivre's Theorem,
\(z^{51}\ =\ (1)^{51}(\cos(51\cdot\frac{2\pi}{3})+i\sin(51\cdot\frac{2\pi}{3}))\\~\\ z^{51}\ =\ (1)^{51}((1)+i(0))\\~\\ z^{51}\ =\ (1)^{51}\\~\\ z^{51}\ =\ 1\)
And so...
\(-z^{51}\ =\ -1\)
Check
+9466 
