Question

I know it has something to do with Unit Circle. But still make me understand this

( Question Attached)

Answer 1

your argument is wrong

Answer 2

while it is true that for \(a+bi\) the arguement \(\theta\) satisfies \(\tan(\theta)=\frac{b}{a}\) it is not necessarily true that \(\theta=\tan^{-1}(\frac{b}{a})\) it depends on what quadrant you are in

Answer 3

in terms of basic angle wouldnt it be tan-1 b/a?

Answer 4

arctan has range restricted to \([-\frac{\pi}{2},\frac{\pi}{2}]\) i.e. quadrants 1 and 4
but our number \(-1-\sqrt{3}i\) is in quadrant 3

Answer 5

thats the third quadrant and the required angle would be pie + pie/3 ?

Answer 6

you need to see what quadrant you are in for the angle, becuase as i said arctangent only does it if you are in quadrants 1 or 4.

Answer 7

of course the angle is not unique, there are infinitely many angles with the same sine and cosine.
so you could say \(-\frac{2\pi}{3}\) or \(\frac{4\pi}{3}\) and you would be right

Answer 8

sometimes a restriction is given say the angle has to be between 0 and \(2\pi\) or
\(-\pi\) and \(\pi\)

Answer 9

but agian if you are in quadrants 2 or 3 arctangent will not be the right answer

Answer 10

in this case there is no range mentioned.

Answer 11

than you can pick anything that looks like \(-\frac{2\pi}{3}+2k\pi\)