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Thread: Polar form of -1 + i.

  1. #1
    Newbie Critter314's Avatar
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    Polar form of -1 + i.

    Write $\displaystyle z=i-1$ in form of $\displaystyle {re}^{iv}$
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  2. #2
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    It's a math exercise! And it is asking you to change a complex number written in Cartesian form to its "polar form". The simplest way to do that is to graph i- 1 as (-1, 1) (using the x-axis as the real axis, the y-axis as the imaginary axis). Now, your "r" is the distance from (0, 0) to (-1, 1) and your "v" is the angle the line from (0,0) to (-1, 1) makes with the positive real (x) axis. Can you calculate those?
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  3. #3
    Newbie Critter314's Avatar
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    Well $\displaystyle r=\sqrt{1^2 + 1^2}=\sqrt{2}$
    Angle is $\displaystyle 315$
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    r is $\displaystyle \sqrt{1^2+1^2} = \sqrt{2}$

    While $\displaystyle \theta = \tan^{-1}\left(\frac{-1}{1}\right), \theta \in (-\pi, \pi)$
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  5. #5
    Newbie Critter314's Avatar
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    I should type answer in radians?
    Then $\displaystyle z={e\sqrt{2}}^{(i*{-0.758})}$ ?
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    I get $\displaystyle \displaystyle z= \sqrt{2}e^{\frac{3\pi}{4}i} $
    Last edited by pickslides; Mar 22nd 2011 at 12:46 PM. Reason: bad latex.
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  7. #7
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    Quote Originally Posted by Critter314 View Post
    Write $\displaystyle z=i-1$ in form of $\displaystyle {re}^{iv}$
    $\displaystyle \text{Arg}(-1+\imath)=\dfrac{3\pi}{4}$.

    So $\displaystyle \sqrt{2}\exp\left(\dfrac{3\imath\pi}{4}\right).$
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  8. #8
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    How do I find an argument?
    315 degrees in radians is $\displaystyle \frac{7\pi}{4}$
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  9. #9
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    Quote Originally Posted by Critter314 View Post
    How do I find an argument?
    The principle value of the argument of a complex number $\displaystyle z=a+bi$ not on any axis is found by the following.
    $\displaystyle Arg(z) = \left\{ {\begin{array}{rl}
    {\arctan \left( {\frac{b}
    {a}} \right),} & {a > 0} \\
    {\arctan \left( {\frac{b}
    {a}} \right) + \pi ,} & {a < 0\;\& \,b > 0} \\ \\
    {\arctan \left( {\frac{b}
    {a}} \right) - \pi ,} & {a < 0\;\& \,b < \pi } \\
    \end{array} } \right.$

    Please note that $\displaystyle \mathif{i}-1=-1+\mathif{i}$ and there $\displaystyle a=-1~\&~b=1$.
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  10. #10
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    Quote Originally Posted by Critter314 View Post
    How do I find an argument?
    315 degrees in radians is $\displaystyle \frac{7\pi}{4}$
    315= 7(45) which is just 45 degrees short of the full circle. That would correspond to 1- i, not -1+ i.
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