Hi Deepak,

$\large \frac{1}{2^i}$ can be written $2^{-i}$ and Euler's formula says that for a real number $\theta$

\[e^{i \theta} =cos \theta + i \sin \theta\]

so you can simplify

\[z = \frac{1}{2^i}\]

if you can write $z$ in the form $e^{i \theta}.$

Since $z = 2^{-i}$ you can take the natural logarithm of each side to get

\[log(z) = \log\left(2^{-i}\right) = -i \log(2)\]

and applying the exponential function to each side to get

\[z = e^{-i \log(2)} = e^{i[-\log(2)]}.\]

Penny