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

Mar 10, 2017

The imaginary component is

$\sin \left(2 \log \left(2 \sqrt{2}\right)\right) {e}^{2 \phi}$ with $\phi = \arctan \left(\frac{1 + \sqrt{3}}{1 - \sqrt{3}}\right)$

#### Explanation:

$\frac{1 + i}{1 - i \sqrt{3}} = \left(1 + i\right) \frac{1 + i \sqrt{3}}{4} = \frac{1}{4} \left(1 - \sqrt{3} + i \left(1 + \sqrt{3}\right)\right)$

Now we will reduce this complex number to the exponential form by making

$x + i y = \sqrt{{x}^{2} + {y}^{2}} {e}^{i \phi}$ where $\phi = \arctan \left(\frac{y}{x}\right)$

so making $x = 1 - \sqrt{3}$ and $y = 1 + \sqrt{3}$ we have

$\sqrt{{x}^{2} + {y}^{2}} = 2 \sqrt{2}$ and $\phi = \arctan \left(\frac{1 + \sqrt{3}}{1 - \sqrt{3}}\right)$

now proceeding

${\left(\frac{1 + i}{1 - i \sqrt{3}}\right)}^{- 2 i} = {\left(2 \sqrt{2}\right)}^{- 2 i} {\left({e}^{i \phi}\right)}^{- 2 i} = {\left({e}^{\lambda}\right)}^{- 2 i} {\left({e}^{i \phi}\right)}^{- 2 i} = {e}^{- 2 i \lambda} {e}^{2 \phi}$

Here $\lambda$ is such that ${e}^{\lambda} = 2 \sqrt{2}$ or $\lambda = \log \left(2 \sqrt{2}\right)$

Finally, using de Moivre's identity

${e}^{i x} = \cos x + i \sin x$ we arrive at

${\left(\frac{1 + i}{1 - i \sqrt{3}}\right)}^{- 2 i} = \left(\cos \left(2 \log \left(2 \sqrt{2}\right)\right) + i \sin \left(2 \log \left(2 \sqrt{2}\right)\right)\right) {e}^{2 \phi}$

and the imaginary component is

$\sin \left(2 \log \left(2 \sqrt{2}\right)\right) {e}^{2 \phi}$ with $\phi = \arctan \left(\frac{1 + \sqrt{3}}{1 - \sqrt{3}}\right)$