- #1
nacho-man
- 171
- 0
another question:
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.
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.