Csdrhrt

I'm having a problem with following equation:

I'm applying $(a+b)^2$ and $(a-b)^2$, but am unable to get the correct answer.

$$(sqrt{3}+i)^{2017} + (sqrt{3} - i)^{2017}$$

I would use the Euler's formula. Say $z = sqrt{3} + i$.

So $tan(phi) = 1/sqrt{3}$

and $phi = pi/6$

and $r = |z|= sqrt{1^2 + sqrt{3}^2} = sqrt{4} = 2$

Meaning $z = 2 cdot e^{i pi/6}$, $overline z = 2 cdot e^{-i pi/6}$

$z^{2017} = 2^2017 cdot e^{i cdot 2017 cdotpi/6}$, $(overline z)^{2017} = 2^{2017} cdot e^{-i cdot 2017 cdotpi/6}$

Converting this back from $w= |w| cdot e^{i cdot psi}$ to the form $w = a + ib$
with $Re(w) = a = |w|cos(psi)$ and $Im(w) = b = |w|sin(psi)$

gives us
$$z^{2017} + (overline z)^{2017} = \
Re(z^{2017}) + Re((overline z)^{2017}) + icdot (Im(z^{2017}) + Im((overline z)^{2017})) = \
2^{2017} cos(2017 cdot pi / 6) + 2^{2017} cos(- 2017 cdot pi /6) + icdot (2^{2017} sin(2017 cdot pi / 6) + 2^{2017} sin(- 2017 cdot pi / 6)) = \
2^{2017} cos(2017 cdot pi / 6) + 2^{2017} cos(2017 cdot pi /6) + icdot (2^{2017} sin(2017 cdot pi / 6) - 2^{2017} sin(2017 cdot pi / 6)) = \
2^{2017} (2 cdot cos(2017 cdot pi /6)) = \
2^{2017} (2 cdot cos(pi /6)) = \
2^{2017} (2 frac{sqrt{3}}{2}) = \
2^{2017} sqrt{3}
$$

Hint:

It's probably easier to write $sqrt 3+i=re^{itheta}$ and $sqrt 3-i=re^{-itheta}$, by finding $r,theta$, and then use $$rcos(theta)=rfrac{e^{itheta}+e^{-itheta}}{2}$$

Let $z=(sqrt 3 +i)^{2017}$, then $$z=2^{2017}e^{2017 ipi /6}$$

$$Re(z)=2^{2017}cos {2017pi /6} =2^{2017} cos {pi/6}$$

$z+bar z =2Re(z)=2^{2017}sqrt 3$