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Math Help - Modulus and Argument

  1. #1
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    Modulus and Argument

    Find modulus and argument of: - 2 - \sqrt {12} i

    \left| z \right| = \sqrt {( - 2)^2  + ( - \sqrt {12} )^2 }  = 4

    <br />
\tan (Argz) = \frac{y}<br />
{x}<br />

    Argz = \tan ^{ - 1} \frac{y}<br />
{x} = \tan ^{ - 1} \frac{{ - \sqrt {12} }}<br />
{{ - 2}} = \tan ^{ - 1} \sqrt 3  = \frac{\pi }<br />
{6}<br />

    The correct answer is:  - \frac{{2\pi }}<br />
{3}

    Could someone explain to me why I have to subtract \pi from my final answer.

    Thanks
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by r_maths View Post
    Find modulus and argument of: - 2 - \sqrt {12} i

    \left| z \right| = \sqrt {( - 2)^2  + ( - \sqrt {12} )^2 }  = 4

    <br />
\tan (Argz) = \frac{y}<br />
{x}<br />

    Argz = \tan ^{ - 1} \frac{y}<br />
{x} = \tan ^{ - 1} \frac{{ - \sqrt {12} }}<br />
{{ - 2}} = \tan ^{ - 1} \sqrt 3  = \frac{\pi }<br />
{6}<br />

    The correct answer is:  - \frac{{2\pi }}<br />
{3}

    Could someone explain to me why I have to subtract \pi from my final answer.

    Thanks
    I don't like using the arctan function. It makes problems because of the restrictions of the domain ><

    z=4 \left(-\frac 12- i \frac{\sqrt{3}}{2}\right)=4(\cos \theta+i \sin \theta)
    Note that the cosine and the sine are both negative. So \theta will be in the third quadrant. That is \theta lies between [3 \pi/2;2 \pi] +2k \pi

    With the specific trig values you should know, \theta=-\frac{2 \pi}{3} +2k \pi


    You also could've written :
    z=-(2+\sqrt{12} i)=e^{i \pi} \left(4 \left(\frac 12+ \frac{\sqrt{12}}{2} i \right)\right)
    Last edited by Moo; December 15th 2008 at 09:12 AM.
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  3. #3
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    Hello, r_maths!

    Find modulus and argument of: -2 -\sqrt{12}i

    \tan (\text{Arg}z) \:= \:\frac{y}{x}

    \text{Arg}z \:= \:\tan^{-1}\left(\frac{-\sqrt{12}}{-2}\right) \:=\: \tan^{-1}\sqrt{3} \:=\: \frac{\pi }{6} . . . . no, you mean: {\color{blue}\frac{\pi}{3}}

    The correct answer is: . -\frac{2\pi}{3} . . . . or {\color{blue}\frac{4\pi}{3}}

    Could someone explain to me why I have to subtract \pi from my final answer.

    Plot the complex number on the Argand diagram.
    Code:
                    |
                    |
                -2  |
          - - + - - + - - - -
              |    /|
           __ |   / |
         -√12 |  /  |
              | /   |
              |/    |
              *     |
                    |

    The argument is an angle in Quadrant 3.

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