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Math Help - Principal Argument of the Complex Number -1 + i sqrt(3)

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    Principal Argument of the Complex Number -1 + i sqrt(3)

    What is the principal argument of the complex number -1+i\sqrt{3}?

    My attempt:

    \begin{array}{rcrcrc}<br />
tan\alpha=\frac{\sqrt{3}}{-1}\\<br />
\alpha=\tan^{-1}(\frac{\sqrt{3}}{-1})\\<br />
=\frac{-1}{3}\pi\\<br />
arg(z)=\pi-\alpha\\<br />
=\frac{4}{3}\pi<br />
\end{array}

    However, I know this is wrong because it does not fit the inequality -\pi<\theta\leq\pi.
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    Quote Originally Posted by Viral View Post
    What is the principal argument of the complex number -1+i\sqrt{3}?

    My attempt:

    \begin{array}{rcrcrc}<br />
tan\alpha=\frac{\sqrt{3}}{-1}\\<br />
\alpha=\tan^{-1}(\frac{\sqrt{3}}{-1})\\<br />
=\frac{-1}{3}\pi\\<br />
arg(z)=\pi-\alpha\\<br />
=\frac{4}{3}\pi<br />
\end{array}

    However, I know this is wrong because it does not fit the inequality -\pi<\theta\leq\pi.

    remember you can add(or subtact) multiples of 2\pi to find a co-terminal angle that falls where you want it

    \frac{4}{3}\pi-2\pi=-\frac{2}{3}\pi
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  3. #3
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    Hmm, I'm not quite sure what you mean. Can you explain it in more details and cite proof?
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    Behold, the power of SARDINES!
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    Quote Originally Posted by Viral View Post
    Hmm, I'm not quite sure what you mean. Can you explain it in more details and cite proof?
    ummm...

    Well lets take an example.

    Say you have the angle

    \frac{\pi}{4} and lets compare that with the angle.

    \frac{9\pi}{4}.

    \frac{9\pi}{4}=2\pi+\frac{\pi}{4}.

    So to get to the angle \frac{9\pi}{4} I would rotate all the way through 2\pi radians (one around the circle so I am back where I started) then go \frac{\pi}{4} radians.

    Moral of the story.

    I can go around any multiple of 2\pi radians as many times as I want and still end up in the same place.
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    Just to make sure, 2\pi is one whole circle?
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    Quote Originally Posted by Viral View Post
    Just to make sure, 2\pi is one whole circle?
    yes that is correct.
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    I don't think this warrants another topic as it's a simple question.

    The modulus argument form is r(cos\theta+isin\theta). However, it could also be r(cos\theta-isin\theta). How do I know which to use?
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  8. #8
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    Quote Originally Posted by Viral View Post
    I don't think this warrants another topic as it's a simple question.

    The modulus argument form is r(cos\theta+isin\theta). However, it could also be r(cos\theta-isin\theta). How do I know which to use?
    r(\cos (-\theta) + i \sin (-\theta)) = r(\cos\theta - \sin\theta). The convention is to use r(\cos\theta+i \sin\theta).
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  9. #9
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    Hmm, so the one I stated first is the only one I can use. This webpage has used both of my examples >< .
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  10. #10
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    If z = a + bi,\,a \ne 0
    \arg (z) = \left\{ {\begin{array}{rl}<br />
   {\arctan \left( {\frac{b}<br />
{a}} \right),} & {a > 0}  \\<br />
   {\arctan \left( {\frac{b}<br />
{a}} \right) + \pi ,} & {a < 0\;\& \,b > 0}  \\<br />
   {\arctan \left( {\frac{b}<br />
{a}} \right) - \pi ,} & {a < 0\;\& \,b < 0}  \\<br />
 \end{array} } \right.
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