How to Convert Complex Numbers from Rectangular to Exponential Form?

[elcotufa]

Homework Statement

Trying to write
$$-8\pi - 8\pi\sqrt3 j$$

in exponential

I got the coefficient as 16 pi

but to get the theta in top of the exponential I have to do

the inverse tangent of $$\frac{-8\pi} {-8\pi\sqrt3 j}$$


I know it is pi over 3, but what is the easiest way to find the multiplicative factor of pi? it is four I just want to know how to find it for other harder examples



answer $$16e^{\frac{j4\pi} 3}$$


Input appreciated

 

[HallsofIvy]

Homework Statement

Trying to write
$$-8\pi - 8\pi\sqrt3 j$$

in exponential

I got the coefficient as 16 pi

but to get the theta in top of the exponential I have to do

the inverse tangent of $$\frac{-8\pi} {-8\pi\sqrt3 j}$$

The "argument", or $\theta$ for x+ jy is arctan(y/x), not arctan(x/y) and certainly not arctan(x/jy)! You want
$$arctan(\frac{-8\pi\sqrt{3}}{-8\pi}= arctan(\sqrt{3})$$
because, of course, the "-8" terms cancel. Now you could use a calculator or, perhaps better, imagine a right triangle with opposite side of length $\sqrt{3}$ and near side 1 (because tan= opposite side/near side). By the Pythagorean theorem, the hypotenuse has length $\sqrt{(\sqrt{3})^2+ 1}= 2$. That is, you are looking for an angle that has sine (opposite side divided by hypotenuse) [tex]\frac{-\sqrt{3}}{2}[/itex] and cosine $$\frac{-1}{2}$$. The negatives are because the real and imaginary parts of the number you give are both - so the point is in the fourth quadrant, not the first.
$sin(\theta)= \sqrt{3}/2$ and $cos(\theta)= 1/2$ should be among the "special angles" you learned long ago.

I know it is pi over 3, but what is the easiest way to find the multiplicative factor of pi? it is four I just want to know how to find it for other harder examples

No, you do NOT know "it is pi over 3" because pi/3 is in the first quadrant and the value you want is in the fourth. 4pi/3 does happen to be in the fourth quadrant.

answer $$16e^{\frac{j4\pi} 3}$$


Input appreciated