Tricky Complex number simplification

[ognik]

Hi - in an example, I can't follow the working from one of the steps to the next, the 2 steps are:

$... \sqrt{\frac{1}{2}\left(1-i\right)} = \sqrt{\frac{1}{\sqrt{2}}{e^{-i(\frac{\pi}{4}-2n\pi)}}}$

I can see they equate $ \frac{1-i}{\sqrt{2}} = e^{-i(\frac{\pi}{4}-2n\pi)}$, and I can see the $ 2n\pi $ allows for n roots (although I don't know why $ - 2n\pi $ instead of $ +2n\pi $?)

But I can't see how $ (1-i) = \sqrt{2}.e^{-i\frac{\pi}{4}} $.

I tried using $ e^{i\frac{\pi}{2}} = i $ and $ e^{i\pi} = -1 $, so $ 1-i = -e^{i\pi} - e^{i\frac{\pi}{2}} = -2e^{i\frac{\pi}{2}} \frac{(e^{i\frac{\pi}{2}}+ 1) }{2} $...trying to get the bracket to $cos\frac{\pi}{2} = 1 $ but I'm stuck there?

 

[Euge]

The modulus of $1 - i$ is $\sqrt{1^2 + (-1)^2} = \sqrt{2}$ and the principal argument is $\arctan((-1)/1) = \arctan(-1) = -\pi/4$. So the polar form of $1 - i$ is $\sqrt{2}e^{-i\pi/4}$.

 

[ognik]

That approach will also be useful to remember, thanks Euge.