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Math Help - Complex numbers: Argument of Z help

  1. #1
    RAz
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    Complex numbers: Argument of Z help

    I'm having some problems with finding, in radians, the argument of z and adding rotations of 180 to keep it in range. I know that I have to use:

    tan=\frac{y}{x}

    and I know exact trigonometric ratios, and the ASTC in the polar plane. But I don't know where Arg Z is positive, and I don't know when to add 2\pi or even just \pi (pi = 180 in radians), to keep Arg Z in the range of -\pi < 0 <\pi. As you can see I am very confused, any help would be great.
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    MHF Contributor Calculus26's Avatar
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    Consider 1 + i and - 1 - i

    In both cases arctan(y/x) = pi/4

    since 1 + i is in the first quadr arg(z) = pi/4

    hoewever for - 1 -i we know this is in the 3d quadrant so we add pi for the arg

    Similarly consider 1 - i and -1 + i

    arctan(y/x) = -pi/4 for -1 + i is in the 2 quadrant so add pi to find argz
    for 1- i is in the fourth quadrant so add 2pi to find argz

    In general locate the quadrant in which z lies to determine arg z
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  3. #3
    Flow Master
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    Quote Originally Posted by RAz View Post
    I'm having some problems with finding, in radians, the argument of z and adding rotations of 180 to keep it in range. I know that I have to use:

    tan=\frac{y}{x}

    and I know exact trigonometric ratios, and the ASTC in the polar plane. But I don't know where Arg Z is positive, and I don't know when to add 2\pi or even just \pi (pi = 180 in radians), to keep Arg Z in the range of -\pi < 0 <\pi. As you can see I am very confused, any help would be great.
    Plot your value of z on an Argand diagram. That makes it very easy to see the ballpark your argument lies in.
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    This function will work.
    <br />
\arg \left( {x + yi} \right) = \left\{ {\begin{array}{ll}<br />
   {\arctan \left( {y/x} \right),} & {x > 0}  \\<br />
   {\arctan \left( {y/x} \right) + \pi ,} & {x < 0\;\& \,y > 0}  \\<br />
   {\arctan \left( {y/x} \right) - \pi ,} & {x < 0\;\& \,y < 0}  \\<br /> <br />
 \end{array} } \right.
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  5. #5
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    Since you want to keep Arg z between -\pi and \pi, Arg z is positive as long as the complex part is postive, negative if the comples part is negative.
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  6. #6
    Senior Member pankaj's Avatar
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    Quote Originally Posted by RAz View Post
    I'm having some problems with finding, in radians, the argument of z and adding rotations of 180 to keep it in range. I know that I have to use:

    tan=\frac{y}{x}

    and I know exact trigonometric ratios, and the ASTC in the polar plane. But I don't know where Arg Z is positive, and I don't know when to add 2\pi or even just \pi (pi = 180 in radians), to keep Arg Z in the range of -\pi < 0 <\pi. As you can see I am very confused, any help would be great.
    First find angle \alpha using \tan\alpha=\left|\frac{y}{x}\right|,where \alpha is an angle lying between 0 and \frac{\pi}{2}

    Now arg(z) is defined as the angle which the line segment joining origin and z makes with positive direction of the real axis.
    Accordingly,
    If x>0,y>0,then arg(z)=\alpha

    If x<0,y>0,then arg(z)=\pi-\alpha

    If x<0,y<0,then arg(z)=\alpha-\pi

    If x>0,y<0,then arg(z)=-\alpha

    If y=0 and x>0,then arg(z)=0,If y=0 and x<0,then arg(z)=\pi

    If x=0 and y>0,then arg(z)=\frac{\pi}{2},If x=0 and y<0,then arg(z)=-\frac{\pi}{2}

    Draw all these cases on the graph and keep the definition of arg(z)(underlined part) foremost in your mind and you will understand immediately.
    Last edited by pankaj; August 6th 2009 at 05:02 PM.
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  7. #7
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    Quote Originally Posted by Plato View Post
    This function will work.
    <br />
\arg \left( {x + yi} \right) = \left\{ {\begin{array}{ll}<br />
   {\arctan \left( {y/x} \right),} & {x > 0}  \\<br />
   {\arctan \left( {y/x} \right) + \pi ,} & {x < 0\;\& \,y > 0}  \\<br />
   {\arctan \left( {y/x} \right) - \pi ,} & {x < 0\;\& \,y < 0}  \\<br />
 \end{array} } \right.
    To the former two posters,
    With the exception x=0 that function answers all of your points.
    What problem do you have with that?
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  8. #8
    Senior Member pankaj's Avatar
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    None whatsoever Plato.

    It is just that I remembered the time when I had got stuck in the same concept and finally came out of it when my teacher had explained it to me in the manner I have written.
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  9. #9
    RAz
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    Thanks for your help everyone.
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