Complex number roots problem

Question

Can anyone help me with this? Please.(fix: the function is multivalued)

$$Z=\left(\frac{1-\sqrt{3}i}{1+\sqrt {3}i}\right)^{10}$$ $$\to Z^{\frac{1}{4}}=?$$

Hey everyone!

I'm an electrical engineering freshman, and I'm new to this website. But something tells me you will see hell of a lot of me around here.

Answer 1

The simple approach would be to simplify $\frac {1-i\sqrt 3}{1+i\sqrt 3}$, multiplying top and bottom by the complex conjugate. This will give you a complex number in standard Cartesian form. Then convert it to polar form. Then apply deMoivres theorem.

The fancy way to do it would be to say: $1+i\sqrt 3 = 2e^{\frac \pi3i}$ and $1-i\sqrt 3 = 2e^{-i\frac \pi3}$

Their ratio is: $e^{\frac {2\pi}{3}i}$

$(e^{\frac {2\pi}{3}i})^{10} =(e^{\frac {20\pi}{3}i}) = (e^{6\pi i})(e^{\frac {2\pi}{3}i}) = (e^{\frac {2\pi}{3}i}) $

Answer 2

Hint: Convert $\left(\frac{1-\sqrt{3}i}{1+\sqrt{3}i} \right)$ to polar coordinate.

Answer 3

Hint: $$Z=\left(\frac{1-i\sqrt{3}}{1+i \sqrt {3}} \cdot \frac{1-i\sqrt{3}}{1-i \sqrt {3}}\right)^{10} = \frac{(1-i\sqrt{3})^{20}}{4^{10}}$$

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[ EDIT  (after some clarifications in the comments under the original question) ]    From:

$$Z=\left(\frac{1-i\sqrt{3}}{2}\right)^{20} = \left(\left(\frac{1-i\sqrt{3}}{2}\right)^{5}\right)^4$$

it is clear that $z = \left(\frac{1-i\sqrt{3}}{2}\right)^{5}$ satisfies $z^4 = Z$, so $z$ is one of the values of $Z^{\frac{1}{4}}$. The other 3 values are $z \cdot \omega_1$, $z \cdot \omega_2$, $z \cdot \omega_3$ where $\omega_k = e^{i\, 2 k \pi / 4}$ $\;|\;\; k=1,2,3$ are the non-unit 4th roots of unity, so in the end the solution set is $\{z, z \cdot i, -z, -z \cdot i\}$.