- Thread starter
- #1

convert $|\frac{1-i}{3}|$ to polar form

i am getting $\frac{\sqrt{2}}{3} e^{\frac{i\pi}{4}}$

but the solutions say:

$e^{\frac{-i\pi}{4}}$

i did

$ x = r\cos(\theta)$ and $y=r\sin(\theta)$

so

$\frac{1}{3} = {\frac{\sqrt{2}}{3}}\cos(\theta)$

$\frac{1}{3} = \cos(\theta)$

And thus $\theta = \frac{\pi}{4}$

similarly, the same was determined for $\sin(\theta)$

What did I do wrong, why didn't i obtain a negative?

Also, is there a shorter method to find the ans? I would love if someone could give me a fully worked solution with the most efficient way to get the answer.

THanks.