# Thread: Complex Analysis - Points of Discontinuity

1. ## Complex Analysis - Points of Discontinuity

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

I have no idea what to do with this problem. Help!

2. ## Re: Complex Analysis - Points of Discontinuity

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

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

\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?

3. ## Re: Complex Analysis - Points of Discontinuity

I can't make the leap from

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

$=arctan \frac{2xy}{x^2-y^2}+n\pi$ where $n \in \mathbb{Z}$

4. ## Re: Complex Analysis - Points of Discontinuity

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

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

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