# Thread: Complex Numbers: Cartesian to Polar

1. ## Complex Numbers: Cartesian to Polar

Hi there,

If I have a complex number z = 1 + root(3)i

Then changing it to polar form is z = r cis(11pi/6)

But my book does mention that the angle must lie between –pi and pi.

Using logic from circular trigonometry 11pi/6 = -pi/6

Therefore in polar form z = r cis(-pi/6) and not r cis(11pi/6) ?

What does this restriction on the angle represent?

2. Most likely just restricting the domain of the Cosine function.

3. Originally Posted by Bushy
Hi there,

If I have a complex number z = 1 + root(3)i

Then changing it to polar form is z = r cis(11pi/6)

But my book does mention that the angle must lie between –pi and pi.

Using logic from circular trigonometry 11pi/6 = -pi/6

Therefore in polar form z = r cis(-pi/6) and not r cis(11pi/6) ?

What does this restriction on the angle represent?
How could $\theta = \frac{11\pi}{6}$ when $z$ is clearly in the first quadrant?

Remember that $r = \sqrt{1^2 + (\sqrt{3})^2} = 2$

and

$\theta = \arctan{\frac{\sqrt{3}}{1}} = \frac{\pi}{3}$.

Therefore $z = 2\,\textrm{cis}\,\frac{\pi}{3}$.

4. Originally Posted by Prove It
How could $\theta = \frac{11\pi}{6}$ when $z$ is clearly in the first quadrant?

Sorry I meant to write z = 1-root(3)i

Then....

changing it to polar form is z = r cis(11pi/6)

But my book does mention that the angle must lie between –pi and pi.

Using logic from circular trigonometry 11pi/6 = -pi/6

Therefore in polar form z = r cis(-pi/6) and not r cis(11pi/6) ?

What does this restriction on the angle represent?

5. Originally Posted by Bushy
Sorry I meant to write z = 1-root(3)i

Then....

changing it to polar form is z = r cis(11pi/6)

But my book does mention that the angle must lie between –pi and pi.

Using logic from circular trigonometry 11pi/6 = -pi/6

Therefore in polar form z = r cis(-pi/6) and not r cis(11pi/6) ?

What does this restriction on the angle represent?
The reason there is a restriction on the argument is because an INFINITE number of angles actually fit.

Namely $-\frac{\pi}{6} + 2\pi n$, where n is an integer representing the number of times you have gone around the unit circle.

In order to avoid confusion, one had to define the "Principle Argument".

And it was chosen that the restriction would be $-\pi < \theta \leq \pi$.

There is no reason why $0 \leq \theta < 2\pi$ could not have been chosen, but it wasn't, so we have to deal with what we have been given to use.

Anyway, to answer your question, there is a restriction on $\theta$ so that you only get one value of $\theta$ instead of a family of values.