# How to find the Im: z, when z=((1+i)/(1-i*sqrt(3)))^(-2i)?

Jun 30, 2017

$z = {e}^{\frac{7 \pi}{6}} \left(\cos \left(\ln \left(\frac{1}{2}\right)\right) - i \sin \left(\ln \left(\frac{1}{2}\right)\right)\right)$

#### Explanation:

In polar form $1 + i$ can be written as $\sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)$ or $\sqrt{2} \times {e}^{i \frac{\pi}{4}}$

and $1 - i \sqrt{3}$ can be written as $2 \left(\cos \left(- \frac{\pi}{3}\right) + i \sin \left(- \frac{\pi}{3}\right)\right)$ or $2 \times {e}^{- i \frac{\pi}{3}}$

Hence $z = {\left(\frac{1 + i}{1 - i \sqrt{3}}\right)}^{- 2 i}$

= ${\left(\frac{\sqrt{2} \times {e}^{i \frac{\pi}{4}}}{2 \times {e}^{- i \frac{\pi}{3}}}\right)}^{- 2 i}$

= ${\left(\frac{1}{\sqrt{2}} \times {e}^{i \left(\frac{\pi}{4} + \frac{\pi}{3}\right)}\right)}^{- 2 i}$

= ${\left(\frac{1}{\sqrt{2}}\right)}^{- 2 i} \times {e}^{- 2 {i}^{2} \frac{7 \pi}{12}}$

= ${\left({e}^{\ln \left(\frac{1}{\sqrt{2}}\right)}\right)}^{- 2 i} \times {e}^{2 \cdot \frac{7 \pi}{12}}$

= e^((7pi)/6)e^(i(-2ln(1/sqrt2))

= ${e}^{\frac{7 \pi}{6}} \left(\cos \left(- 2 \ln \left(\frac{1}{\sqrt{2}}\right)\right) + i \sin \left(- 2 \ln \left(\frac{1}{\sqrt{2}}\right)\right)\right)$

= ${e}^{\frac{7 \pi}{6}} \left(\cos \left(2 \ln \left(\frac{1}{\sqrt{2}}\right)\right) - i \sin \left(2 \ln \left(\frac{1}{\sqrt{2}}\right)\right)\right)$

= ${e}^{\frac{7 \pi}{6}} \left(\cos \left(\ln \left(\frac{1}{2}\right)\right) - i \sin \left(\ln \left(\frac{1}{2}\right)\right)\right)$