- #1

Mayhem

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$$L = \left (\frac{-1+i\sqrt{3}}{2}\right )^6+\left (\frac{-1-i\sqrt{3}}{2}\right )^6+\left (\frac{-1+i\sqrt{3}}{2}\right )^5+\left (\frac{-1-i\sqrt{3}}{2}\right )^5$$

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?