Problem with Complex numbers

MuhTheKuh

I have two small problems with some Questions about Complex numbers I have been looking through

For the 1st one. I do not understand how they got to the value of 4/3*Pi

And secondly I do not understand why in one example he says that its cis(0+...), while in all examples I have done before that one I was told to use cis(Pi+...).

Random Variable

$$\displaystyle \tan \theta = \frac{y}{x} = \frac{ 8 \sqrt{3}}{8} = \sqrt{3}$$

$$\displaystyle \theta = \arctan \sqrt{3} = \frac{\pi}{3}$$

but $$\displaystyle \theta$$ is in the third quadrant since both x and y are negative

so $$\displaystyle \theta = \frac{4 \pi}{3}$$

EDIT: For the second one, all real positive numbers are on the positive x axis. So in that case $$\displaystyle \theta = 0$$

Last edited:

MuhTheKuh

But why is it in the second example with arctan(0-/81) a Pi?
And since its Pi/3 and in the 3rd quadrant shouldnt it be -2/3*Pi?

HallsofIvy

MHF Helper
The point is that we are thinking of the complex number "a+ bi" as a point in the xy-plane. a+ bi corresponds to the point (a, b). In polar coordinates, since $$\displaystyle x= r cos(\theta)$$ and $$\displaystyle y= r sin(\theta)$$, $$\displaystyle \frac{y}{x}= \frac{r sin(\theta)}{r cos(\theta)}= \frac{sin(\theta)}{cos(\theta)}= tan(\theta)$$ so that $$\displaystyle \theta= arctan(\frac{y}{x})$$ as long as x is not 0. Even for x= 0, since tan(\theta) goes to infinity as $$\displaystyle \theta$$ goes to $$\displaystyle \pi/2$$ and to negative infinity as $$\displaystyle \theta$$ goes to $$\displaystyle -\pi/2$$, we have $$\displaystyle \theta= \pi/2$$ or $$\displaystyle \theta= -\pi/2$$.

Since $$\displaystyle \theta$$ is the angle the line from (0, 0) to (a, b) makes with the positive real axis, you should be able to see, without any calculation that (1) for any positive real number, $$\displaystyle \theta= 0$$, (2) for any positive real number times i (like 2i or $$\displaystyle i\sqrt{3}$$) $$\displaystyle \theta= \pi/2$$, (3) for any negative real number $$\displaystyle \theta= \pi$$ (or $$\displaystyle -\pi$$ since we can add or subtract any multiple of $$\displaystyle 2\pi$$ and have the same position), (4) for any negative real number times i (like -2i or $$\displaystyle -i\sqrt{3}$$) $$\displaystyle \theta= 2\pi/2$$ (or $$\displaystyle -\pi/2$$).

Random Variable

But why is it in the second example with arctan(0-/81) a Pi?
And since its Pi/3 and in the 3rd quadrant shouldnt it be -2/3*Pi?
because all negative real numbers are on the negative x-axis

and $$\displaystyle \frac{4 \pi}{3} = -\frac{2 \pi}{3}$$

Random Variable

I incorrectly stated the reasoning for why the angle in the second problem is $$\displaystyle 0$$. Check my "EDIT".