# Thread: Need help with polar form

1. ## Need help with polar form

For example let's take

- 1 + sqrt(3) i

I want t change the above expression into polar form.

I know how to change it to polar form in the form of r*cis(thetha)

I know how to get r and thetha, but my issue lies with getting the exact thetha value (meaning I want the answer in a fraction format with pi and not in decimal format as the calculator gives out)

So for the initial expression above, the exact value of thetha would be 2*pi/3

Could someone explain how to get 2pi/3 ?

Thank you.

2. ## Re: Need help with polar form

$\tan \theta = \dfrac{\sqrt{3}}{-1} = -\sqrt{3}$

You are looking for a value of $\theta$ that makes the statement true. That is when $\cos \theta = -\dfrac{1}{2}$ and $\sin \theta = \dfrac{\sqrt{3}}{2}$ (notice the opposite signs for these, sine is positive while cosine is negative). You know you want a solution in the second quadrant (where sine is positive and cosine is negative). This is because $-1 = r\cos \theta$ with $r$ nonnegative and $\sqrt{3} = r\sin \theta$.

3. ## Re: Need help with polar form

I get the quadrant part, but how do we end up with (2*pi)/3 as the angle?

4. ## Re: Need help with polar form

Originally Posted by gkk
For example let's take - 1 + sqrt(3) i
I want t change the above expression into polar form.

So for the initial expression above, the exact value of thetha would be 2*pi/3
Could someone explain how to get 2pi/3
Suppose that $\displaystyle x\cdot y\ne 0$ then
$\displaystyle \theta= \arg(x + yi) = \left\{ {\begin{array}{{rl}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right.$ note that $-\pi<\theta<\pi.$

In your case $x=-1<0~\&~y=\sqrt3>0$ so $\theta=\pi+\arctan\left(\frac{\sqrt3}{-1}\right)=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

5. ## Re: Need help with polar form

$\arctan (-\sqrt{3})$

6. ## Re: Need help with polar form

Originally Posted by gkk
For example let's take

- 1 + sqrt(3) i

I want t change the above expression into polar form.

I know how to change it to polar form in the form of r*cis(thetha)

I know how to get r and thetha, but my issue lies with getting the exact thetha value (meaning I want the answer in a fraction format with pi and not in decimal format as the calculator gives out)

So for the initial expression above, the exact value of thetha would be 2*pi/3

Could someone explain how to get 2pi/3 ?

Thank you.
What you need to know are the basic triangles For example (click to enlarge):

One of these is the 45-45-90 triangle and you can read from it that the sine and cosine of 45 degrees is $\frac 1 {\sqrt 2}$. The other is an equilateral triangle with side 2 and an altitude of $\sqrt 3$. From it you can read that $\sin 30 = \frac 1 2$ and $\cos 30 = \frac {\sqrt 3} 2$. And presumably you know that $30^\circ = \frac \pi 6$, $60^\circ = \frac \pi 3$, and $45^\circ = \frac \pi 4$ in radians.

Now, for your particular problem, draw a 30-60-90 triangle on your xy coordinate system as follows, noting that the x leg is negative, like so:

Now, you know that $60^\circ$ angle in the second quadrant is $\frac \pi 3$. Subtract that from $\pi$ to get your polar angle.
You can use this method to get all the multiples of 45 and 30 degrees as radians. Hope that helps.

7. ## Re: Need help with polar form

Thanks for showing the rules. It sure helps.

8. ## Re: Need help with polar form

Walagaster, this was the explanation I was looking for. I understand now. Thank you so much!!!