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Thread: Need help with polar form

  1. #1
    gkk
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    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.
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  2. #2
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    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 $\sqrt{3} = r\sin \theta$.
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  3. #3
    gkk
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    Re: Need help with polar form

    I get the quadrant part, but how do we end up with (2*pi)/3 as the angle?
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  4. #4
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    Re: Need help with polar form

    Quote Originally Posted by gkk View Post
    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 $\theta=\pi+\arctan\left(\frac{\sqrt3}{-1}\right)=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$
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  5. #5
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    Re: Need help with polar form

    $\arctan (-\sqrt{3}) $
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  6. #6
    Junior Member Walagaster's Avatar
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    Re: Need help with polar form

    Quote Originally Posted by gkk View Post
    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):

    Need help with polar form-triangles.jpg
    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:
    Need help with polar form-polarcoordinate.jpg
    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.
    Last edited by Walagaster; Jun 13th 2018 at 09:44 AM.
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  7. #7
    gkk
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    Re: Need help with polar form

    Thanks for showing the rules. It sure helps.
    Last edited by gkk; Jun 13th 2018 at 09:50 AM.
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  8. #8
    gkk
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    Re: Need help with polar form

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