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Thread: Complex Numbers to polar..

  1. #1
    Junior Member
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    Complex Numbers to polar..

    Could use some help on this... as i remember it seemed pretty easy, but doesnt it always?

    $\displaystyle 1 - \sqrt 3i \longrightarrow polar $

    $\displaystyle \sqrt 5 + i \longrightarrow polar $

    $\displaystyle 2(cos (\pi/6) + sin (\pi/6i) \longrightarrow rect $

    some direction would be helpful. thanks...
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  2. #2
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    Quote Originally Posted by cjmac87 View Post
    Could use some help on this... as i remember it seemed pretty easy, but doesnt it always?

    $\displaystyle 1 - \sqrt 3i \longrightarrow polar $

    $\displaystyle \sqrt 5 + i \longrightarrow polar $

    $\displaystyle 2(cos (\pi/6) + sin (\pi/6i) \longrightarrow rect $

    some direction would be helpful. thanks...
    If you have a complex number z in the form of a +bi, then the polar conversion is $\displaystyle z = r(\cos( \theta ) + i\sin( \theta))$ and $\displaystyle a=r\cos( \theta )$ and $\displaystyle b=r\sin( \theta )$, where $\displaystyle r=\sqrt{a^2+b^2}$. If you draw out a point on the complex plane and work with the right triangle it makes, you can derive these or verify them easily.
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