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Math Help - [SOLVED] Help determining principal Arg

  1. #1
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    [SOLVED] Help determining principal Arg

    Can anyone show me how to determine the Arg(2+i) ?

    I know that \theta=sin^{-1}(\frac{1}{\sqrt{5}})

    But I don't know how to reduce that any further to determine what theta is...

    Any help?
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  2. #2
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    \text{Arg}(x + yi) = \left\{ {\begin{array}{rl}<br />
   {\arctan \left( {\frac{y}<br />
{x}} \right),} & {x > 0}  \\<br />
   {\arctan \left( {\frac{y}<br />
{x}} \right) + \pi ,} & {x < 0\,\& \,y > 0}  \\<br />
   {\arctan \left( {\frac{y}<br />
{x}} \right) - \pi ,} & {x < 0\,\& \,y < 0}  \\<br /> <br />
 \end{array} } \right.
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  3. #3
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    Sorry, I should've put the whole problem on here. The problem states:

    "By writing the individual factors on the left in exponential form, performing the needed operations, and finally changing back to rectangular coordinates, show that
    \frac{5i}{2+i}=1+2i

    This can be easily shown by just multiplying top and bottom by the conjugate of the denominator, but I have to do it how the directions say. I know how to convert 5i into exponential form, but converting 2+i into exponential form has proven to be more problematic since arctan(1/2) isn't very simple.

    Any help?
    Last edited by paupsers; January 14th 2010 at 02:35 PM.
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  4. #4
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    Quote Originally Posted by paupsers View Post
    Sorry, I should've put the whole problem on here. The problem states:

    "By writing the individual factors on the left in exponential form, performing the needed operations, and finally changing back to rectangular coordinates, show that
    \frac{5i}{2+i}=1+2i"

    This can be easily shown by just multiplying top and bottom by the conjugate of the denominator, but I have to do it how the directions say. I know how to convert 5i into exponential form, but converting 2+i into exponential form has proven to be more problematic since arctan(1/2) isn't very simple.
    You should always post the entire question.

    Let \phi  = \arctan \left( {\frac{1}{2}} \right) then <br />
2 + i = \sqrt 5 e^{i\phi }
    Now do it.
    Last edited by Plato; January 14th 2010 at 02:42 PM. Reason: I missed the 5i
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  5. #5
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    Quote Originally Posted by Plato View Post
    You should always post the entire question.

    There is a mistake in you question: it should be \frac{5i}{2+i}=2{\color{red}-}i
    Let \phi  = \arctan \left( {\frac{1}{2}} \right) then <br />
2 + i = \sqrt 5 e^{i\phi }
    Now do it.
    Actually, the way I posted the question is correct.
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  6. #6
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    Quote Originally Posted by paupsers View Post
    Actually, the way I posted the question is correct.
    You are correct, I missed the 5i

    Anyway,it still works.
    5i=5e^{\frac{i\pi}{2}} so \frac{{5i}}{{2 + i}} = \sqrt 5 e^{\frac{i\pi}{2} } \left( {e^{ - i\phi } } \right) = \sqrt 5 e^{i\left( {\frac{\pi}{2}  - \phi } \right)} .
    Last edited by Plato; January 14th 2010 at 03:36 PM.
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  7. #7
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    Quote Originally Posted by Plato View Post
    You are correct, I missed the 5i

    Anyway,it still works.
    5i=5e^{\frac{i\pi}{2}} so \frac{{5i}}{{2 + i}} = \sqrt 5 e^{\frac{i\pi}{2} } \left( {e^{ - i\phi } } \right) = \sqrt 5 e^{i\left( {\frac{\pi}{2}  - \phi } \right)} .
    This is how far I had gotten before. This is where I don't understand how to get it back into rectangular coordinates.
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  8. #8
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    Nevermind! I figured it out!
    Thanks!
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