Forgetting my inverse tangent, polar form of compelx number ASTC

[thomas49th]

Homework Statement

Find the polar form of 2i − 1


Finding polar form is easy r(cosx + isinx)

call the real part a and imaginary part b

r = sqrt(a+b)

theta = arctan (-2) = - 63.43

This is the wrong angle for theta as it's 116.57 (which is 180 - 64.43), and I guess this if form that quadrant thing

S | A
___|___
|
T | C


okay not the best grid, but can anyone refresh my memory as how to get the correct value of theta?

Thanks :)

 

[tiny-tim]

Hi thomas49th!

(have a theta: θ and a square-root: √ )

(and you meant √(a² + b²))

Yes, your table shows what's positive (sin, tan, cos, or all) …

(it works because cos*tan = sin, so an even number must be negative, so either 1 or all 3 must be positive )

so arctan(-2) can be either top left or bottom right, you can't tell which by looking at it …

you have to go back to the original 2i - 1, which is at (-1,2), so it's obviously top left!