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Thread: Convert into polar form of e raise to power i(theta), please help.

  1. #1
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    Convert into polar form of e raise to power i(theta), please help.

    i1/2

    and

    z1/n



    and

    -9


    where z=x+i y
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  2. #2
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    Re: Convert into polar form of e raise to power i(theta), please help.

    Quote Originally Posted by szak1592 View Post
    i1/2

    and

    z1/n



    and

    -9


    where z=x+i y
    $\displaystyle \displaystyle \begin{align*} i^{\frac{1}{2}} &= \left( e^{i \frac{\pi}{2}} \right)^{\frac{1}{2}} \\ &= e^{i \frac{\pi}{4}} \end{align*}$

    and

    $\displaystyle \displaystyle \begin{align*} z^{\frac{1}{n}} &= \left( r\, e^{i\theta} \right)^{\frac{1}{n}} \\ &= r^{\frac{1}{n}}\,e^{i\frac{\theta}{n}} \end{align*}$

    and

    $\displaystyle \displaystyle -9 = 9\,e^{ i \pi } $
    Last edited by Prove It; Mar 20th 2013 at 02:46 AM.
    Thanks from szak1592
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  3. #3
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    Re: Convert into polar form of e raise to power i(theta), please help.

    For general $\displaystyle x+ iy= re^{i\theta}$ with $\displaystyle r= \sqrt{x^2+ y^2}$, $\displaystyle \theta= arctan(y/x)$
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  4. #4
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    Re: Convert into polar form of e raise to power i(theta), please help.

    Quote Originally Posted by HallsofIvy View Post
    For general $\displaystyle x+ iy= re^{i\theta}$ with $\displaystyle r= \sqrt{x^2+ y^2}$, $\displaystyle \theta= arctan(y/x)$
    $\displaystyle \displaystyle \theta$ is only $\displaystyle \displaystyle \arctan{ \left( \frac{y}{x} \right) }$ if the complex number is in the first quadrant. Otherwise you need to evaluate the angle in the first quadrant and then transform.
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