Writing $((i-\sqrt3 )/(i-1))^{24}$ in algebrical form

Question

$$ \left(\frac{i-\sqrt3 }{i-1}\right)^{24} $$ is to be written in algebrical form. I'd like to forget about the exponent for the moment, and focus on the base. If I'm right it can be rewritten like: $$ \left(\frac{1+\sqrt3}{2}\right)+i\left(\frac{\sqrt3-1}{2}\right) $$ Any hint on how to go on? I'm trying to write it in polar form, but without success.

Answer 1

It is better to use the following observations:

1. We have $$i-\sqrt{3}=2\left(-\frac{\sqrt{3}}{2}+\frac{i}{2}\right)=2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)=2e^{5i\pi/6}.$$

2. Similarly, $$i-1=\sqrt{2}\,e^{3i\pi/4}.$$

Therefore: \begin{align} \left(\frac{i-\sqrt{3}}{i-1}\right)^{24}=\left[\sqrt{2}e^{i\pi\left(\frac{5}{6}-\frac34\right)}\right]^{24}=2^{12} e^{2\pi i}=4096. \end{align}

Answer 2

Hint: $$\frac{i-\sqrt3 }{i-1} = (i+1)\left(\cos \left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right) = \sqrt{2} \left(\cos \left(\frac{\pi}{4}-\frac{\pi}{6}\right)+i\sin \left(\frac{\pi}{4}-\frac{\pi}{6}\right)\right).$$

Answer 3

HINT:

$$\frac{i-\sqrt3}{i-1}=\frac{\sqrt3-i}{1-i}$$

Now using Euler Formula,

$\sqrt3-i=2\{\cos(-\frac\pi6)+i\sin(-\frac\pi6)\}=2e^{-i\frac\pi6}$

$1-i=\sqrt2\{\cos(-\frac\pi4)+i\sin(-\frac\pi4)\}=\sqrt2e^{-i\frac\pi4}$

We can prove using Euler Formula that $e^{2n\pi i}=1$ where $n$ is an integer

Answer 4

You can write $\frac{1+\sqrt{3}}{2}$ as $cos(\frac{\pi}{3})$+$sin(\frac{\pi}{3})$ and $\frac{\sqrt{3}-1}{2}$ as $cos(\frac{\pi}{3})$-$sin(\frac{\pi}{3})$. Then $cos(\frac{\pi}{3})$+$sin(\frac{\pi}{3})$ $+$ $i$ [$cos(\frac{\pi}{3})$-$sin(\frac{\pi}{3})$ ] will give $cos(\frac{\pi}{3})$ - $i$ $sin(\frac{\pi}{3})$ + $i$[$cos(\frac{\pi}{3})$ - $i$ $sin(\frac{\pi}{3})$] which gives $e^{-{\frac{\pi}{3}}}$[1+$i$] which is $e^{-{\frac{\pi}{3}}}$ $\sqrt{2}e^{\frac{\pi}{4}}$ . Hope you can take it from here

Answer 5

Calculate the square of the revised form you calculated and you will see that you get $\sqrt{3} + i$. Expressing this in modulus/argument form is simple, and so then is raising it to the 12th power.