Complex Addition: $(i+\sqrt3/2)^{200} + (i-\sqrt3/2)^{200}$

Question

Solving a question, I need to find the value of following in between the solution.

$$\left(\frac{i+\sqrt3}{2}\right)^{200} + \left(\frac{i-\sqrt3}{2}\right)^{200}$$

The only useful thing I got was

$$\left(\frac{i+\sqrt3}{2}\right)^{100}\left(\frac{i-\sqrt3}{2}\right)^{100} = 1$$

Which might be useful to complete the square.

Answer 1

$(\frac{i+\sqrt3}{2})$ = $e^{i\theta}$ where $\theta = 30^o$ Now, $$(\frac{i+\sqrt3}{2})^{200} = e^{i200\theta} = (\frac{-1-i\sqrt3}{2})$$
Similarly,$$(\frac{i-\sqrt3}{2})^{200} = e^{i200\theta} = (\frac{-1+i\sqrt3}{2})$$ here $\theta$ being $150^o$ . Now, just sum them up to get $-1$

Answer 2

Actually, we can even generalize:$$z^n+\bar{z}^n=r^ne^{in\theta}+r^ne^{-in\theta}=r^n\left[\cos(n\theta)+i\sin(n\theta)+\cos(-n\theta)+i\sin(-n\theta)\right]$$Using parity of $\sin,\cos$ function we eventually get$$z^n+\bar{z}^n=2r^n\cos(n\theta)$$

In this case $\displaystyle z=\frac{\sqrt{3}}{2}+\frac{1}{2}i\Rightarrow r=1,\theta=\frac{\pi}{6}$ and $n=200$ hence $$ z^n+\bar{z}^n=2\cdot{1^{200}}\cos\left(\frac{4\pi}{3}\right)=-1$$

• I used the fact that $\cos$ is $2\pi$ periodic, hence $\displaystyle \cos\left(\frac{200\pi}{6}\right)=\cos\left(\frac{4\pi}{3}\right)=-\frac{1}{2}$

Answer 3

If $a=\dfrac{i+\sqrt3}2, b=\dfrac{\sqrt3-i}2$

$a+b=\sqrt3,ab=1\implies a^2+b^2=(a+b)^2-2ab=1$

So, $a,b$ are the roots of $t^2-\sqrt3t+1=0$

$\implies(t^2+1)^2=(\sqrt3t)^2$ $\iff t^4-t^2+1=0\implies(t^2+1)(t^4-t^2+1)=0\iff t^6=-1$

$\implies t^{2+6(2n+1)}=(t^6)^{2n+1}\cdot t^2=-t^2$

$\implies a^{12n+8}+b^{12n+8}=-(a^2+b^2)=-1$