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 #2
avatar+9142 
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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 

Nov 21, 2020