High School Computing a tricky complex math problem - where did I go wrong?

[Mayhem]

I stumbled upon the following problem on instagram:
$$L = \left (\frac{-1+i\sqrt{3}}{2}\right )^6+\left (\frac{-1-i\sqrt{3}}{2}\right )^6+\left (\frac{-1+i\sqrt{3}}{2}\right )^5+\left (\frac{-1-i\sqrt{3}}{2}\right )^5$$
The idea is to compute it. Using a calculator, it is supposed to equal 1. My result is something quite different, even though my work seems sound. Where did I go wrong?:

First, we recognize that the denominator of each fraction is really just a complex number in the form $$z = a+bi$$

Thus we get
$$z_1 = a_1+b_1 i = -1+\sqrt{3}i$$
$$z_2 = a_2+b_2 i = -1-\sqrt{3}i$$
by exponentiation and substitution, we may rewrite the equation as
$$\frac{z_1^6}{64}+\frac{z_2^6}{64}+\frac{z_1^5}{32}+\frac{z_2^5}{32}$$
This is still hard to compute due to the exponents, but we can rewrite z_1 and z_2 in exponential form to simplify the problem, for which we will need to determine r and θ for the respective complex numbers.

Plugging in the numbers we get:
$$r_1 = \sqrt{a_1^2+b_1^2} = \sqrt{(-1)^2+\sqrt{3}^2} = 2$$
$$r_2 = \sqrt{a_2^2+b_2^2} = \sqrt{(-1)^2+(-\sqrt{3})^2} = 2$$
and
$$\theta_1 = \arctan{\frac{b_1}{a_1}} = \arctan{\frac{\sqrt{3}}{-1}} = -\frac{\pi}{3}$$
$$\theta_2 = \arctan{\frac{b_2}{a_2}} = \arctan{\frac{\sqrt{-3}}{-1}} = \frac{\pi}{3}$$
We can now rewrite $z_1$ and $z_2$ in exponential form:
$$z_1 = r_1e^{i\theta_1} = 2e^{-i\frac{\pi}{3}}$$
$$z_2 = r_2e^{i\theta_2} = 2e^{i\frac{\pi}{3}}$$
Now we can compute:
$$z_1^6 = 64e^{-i2\pi}$$
$$z_1^5 = 32e^-i\frac{5\pi}{3}$$

$$z_2^6 = 64e^{i2\pi}$$
$$z_2^5 = 32e^i\frac{5\pi}{3}$$
We can plug this back into our equation, and cancelling out the denominators, we get
$$L = e^{-i2\pi}+e^{i2\pi} + e^{-i\frac{5\pi}{3}}+e^{i\frac{5\pi}{3}}$$
Using e^iπa = (-1)^a, we can compute this. It does not equal 1, but some strange expression. I have a strong intuition that I went wrong when calculating the angles, but I'm a bit rusty on this. Could someone help?

 

[etotheipi]

Notice that $-\frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{\frac{2\pi i}{3}}$ and $-\frac{1}{2} - \frac{\sqrt{3}}{2}i = e^{\frac{4\pi i}{3}} = e^{-\frac{2\pi i}{3}}$. You can verify this with Euler's formula. That should make the problem slightly easier.

 

[anuttarasammyak]

As for the RHS of the first L equation, the first and the second terms are conjugate complex.
The third and the fourth terms are conjugate complex.
So L = 2 Re (the first term + the third term).
=2 Re ( e^{6(2/3\ \pi i)}+ e^{5(2/3\ \pi i)} )= 2(1- 1/2 )= 1
which meets your answer.

 

[Delta2]

First of all your result doesn't equal to some strange expression as you say but it is equal to $1+1+0.5+0.5=3$. I think your mistake lies in the choice of the angles, you should take $\theta_1=\frac{2\pi}{3}$ , which still has a $\tan\theta_1=tan\frac{2\pi}{3}=-\sqrt{3}$ and $\theta_2=-\frac{2\pi}{3}$. Why such choice? Because i think when we are looking for the primary argument of a complex number we are looking for a positive argument.

 

[Mayhem]

As for the RHS of the first L equation, the first and the second terms are conjugate complex.
The third and the fourth terms are conjugate complex.
So L = 2 Re (the first term + the third term).
=2 Re ( e^{6(2/3\ \pi i)}+ e^{5(2/3\ \pi i)} )= 2(1- 1/2 )= 1
which meets your answer.

I am not familiar with Re() notation, although I've seen it before. Does it simply mean we only worry about the real portion of the input?

 

[etotheipi]

Yes if $z = a + bi$ then $\text{Re}(z) = a$ and $\text{Im}(z) = b$, i.e. the projections onto the real and imaginary axes respectively.

 

[Mayhem]

Yes if $z = a + bi$ then $\text{Re}(z) = a$ and $\text{Im}(z) = b$, i.e. the projections onto the real and imaginary axes respectively.

It rings a bell now. So, in order to evaluate Re(##re^{i\theta}), one simply has to convert into rectangular form and do as shown?

 

[etotheipi]

It rings a bell now. So, in order to evaluate Re($re^{i\theta}$), one simply has to convert into rectangular form and do as shown?

Sure, that works. If the complex number is $z = re^{i\theta} = r(\cos{\theta} + i\sin{\theta})$ then $\text{Re}(z) = r\cos{\theta}$ and $\text{Im}(z) = r\sin{\theta}$. The geometric interpretation is that the complex number you can think of as a vector in the complex plane and the real/imaginary components are the cosines/sines of the arguments multiplied by the magnitude.