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I can use Mathematica to investigate the continuity of the real-valued function $$\text{Im}\sqrt{x+iy}$$ by drawing a density plot or a plot of the surface.

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Clearly, they are continuous on $R\backslash(-\infty,0]$. However, I am wondering how I can make this argument with pencil and paper. If I write $$\sqrt{x+iy}=\sqrt[4]{x^2+y^2}e^{i\,\text{arg}(x+iy)/2},$$ then it seems I can argue that $$\text{Im}\sqrt{x+iy}=\sqrt[4]{x^2+y^2}\sin\left(\frac12\text{arg}(x+iy)\right)$$ Now, as $x+iy$ approaches the negative real axis from above, then $\text{arg}(x+iy)\rightarrow\pi$ and $\frac12\text{arg}(x+iy)\rightarrow\frac{\pi}{2}$ and $\sin\left(\frac12\text{arg}(x+iy)\right)\rightarrow 1$. Hence: $$\text{Im}\sqrt{x^2+y^2}\rightarrow \sqrt[4]{x^2+y^2}.$$ On the other hand, as $x+iy$ approaches the negative real axis from below, $\text{arg}(x+iy)\rightarrow -\pi$, so $\frac12\text{arg}(x+iy)\rightarrow -\frac{\pi}{2}$ and $\sin\left(\frac12\text{arg}(x+iy)\right)\rightarrow -1$. Hence, $$\text{Im}\sqrt{x^2+y^2}\rightarrow -\sqrt[4]{x^2+y^2}.$$ Hence, the discontinuity on the negative part of the real axis.

My Question: First, I'm wondering if my argument is OK. Second, is there an easier technique, as my next question is to find where the $\text{Im}\sqrt{(x+iy)^3}$ is discontinuous. Again, pretty easy with Mathematica:

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But how to do it with only paper and pencil?

Thanks.

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Your reasoning is correct. An easier way to do it is to write the complex numbers in polar form: $x+y\,i=r\,e^{i\theta}$. Then $$ \sqrt{(x+y\,i)^3}=r^{3/2}e^{\tfrac{3\theta}{2}i}. $$ Since $\sqrt z=\sqrt r\sqrt{e^{i\theta}}=\sqrt re^{i\theta/2}$ has discontinuities in $\theta=\pi+2\,k\,\pi$, or what is the same $\theta/2=\pi/2+k\,\pi$, the discontinuities of $\sqrt{z^3}$ appear when $$ \frac{3\theta}{2}=\frac{\pi}{2}+k\,\pi\quad\text{for integer $k$.} $$ For $k=0$ you get $\theta=\pi/3$, for $k=1$ $\theta=\pi$, and for $k=-1$ $\theta=-\pi/3$.

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  • $\begingroup$ OK, these agree, but why does this work? What is the reasoning behind this computation? $\endgroup$ – David Mar 16 '13 at 17:42
  • $\begingroup$ I have included an explanation in the answer. $\endgroup$ – Julián Aguirre Mar 17 '13 at 16:42

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