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Thread: Complex Analysis - Points of Discontinuity

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    Complex Analysis - Points of Discontinuity

    Find all points of discontinuity of the function $\displaystyle g(z)=Arg(z^2)$.

    I have no idea what to do with this problem. Help!
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    Re: Complex Analysis - Points of Discontinuity

    Quote Originally Posted by tarheelborn View Post
    Find all points of discontinuity of the function $\displaystyle g(z)=Arg(z^2)$.

    I have no idea what to do with this problem. Help!
    You need to write this function as $\displaystyle \displaystyle u(x, y) + i\,v(x, y)$. So

    $\displaystyle \displaystyle \begin{align*} z^2 &= \left(x + y\,i \right)^2 \\ &= x^2 - y^2 + 2xy\,i \\ \arg{\left(z^2\right)} &= \arctan{ \left( \frac{2xy}{x^2 - y^2} \right) } + n\pi \textrm{ where }n \in \mathbf{Z} \end{align*}$

    Where is this discontinuous?
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    Re: Complex Analysis - Points of Discontinuity

    I can't make the leap from

    $\displaystyle =x^2-y^2+2xyi$ to

    $\displaystyle =arctan \frac{2xy}{x^2-y^2}+n\pi$ where $\displaystyle n \in \mathbb{Z}$
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    Re: Complex Analysis - Points of Discontinuity

    Quote Originally Posted by tarheelborn View Post
    I can't make the leap from
    $\displaystyle =x^2-y^2+2xyi$ to
    $\displaystyle =arctan \frac{2xy}{x^2-y^2}+n\pi$ where $\displaystyle n \in \mathbb{Z}$
    If you will note that this question is about the principle value of the argument of a complex number $\displaystyle z=a+bi$ not on any axis is found by the following.
    $\displaystyle Arg(z) = \left\{ {\begin{array}{rl} {\arctan \left( {\frac{b}{a}} \right),} & {a > 0} \\ {\arctan \left( {\frac{b}{a}} \right) + \pi ,} & {a < 0\;\& \,b > 0} \\ \\ {\arctan \left( {\frac{b}{a}} \right) - \pi ,} & {a < 0\;\& \,b < \pi } \\ \end{array} } \right.$

    For any real number $\displaystyle r$ then $\displaystyle Arg(r) = \left\{ {\begin{array}{rl} {0,} & {r \geqslant 0} \\ {\pi ,} & {r < 0} \\ \end{array} } \right.$;
    And $\displaystyle Arg(ri) = \left\{ {\begin{array}{rl} {\frac{\pi }{2},} & {r \geqslant 0} \\ {\frac{{ - \pi }}{2},} & {r < 0} \\ \end{array} } \right.$

    You will need to make adjustments in the above, in that $\displaystyle a=x^2-y^2~\&~b=2xy.$
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