Whether the function $f(z)= \sqrt{xy}$ is analytic at the origin $(0,0)$ or not?
I want to know how to check/verify using Cauchy-Riemann equations.
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Where do you think that function is defined? – zhw. Aug 28 '17 at 19:19
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@zhw.: By choosing appropriate branches, you can define it continuously in the entire complex plane. – hmakholm left over Monica Aug 28 '17 at 19:19
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@HenningMakholm I'm aware of that. But I was asking where the OP thought this function is defined. – zhw. Aug 28 '17 at 19:27
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If it were analytic (rather than just differentiable) at the origin, it would be complex differentiable in an entire neighborhood of the origin. But it is not even real differentiable at $z=\varepsilon+0i$ for any $\varepsilon>0$ (consider for example $f\circ g$ with $g(t)=\varepsilon+it$).
But it can't even be real differentible at the origin. The partial derivatives are all $0$, but on the line $x=y$ we have $f(x+iy)=x$ -- which is not approximated by $0x+0y$.
hmakholm left over Monica
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@zhw, Thank you $$f(x,y)= \begin{cases} &\sqrt{xy} \quad\text{for } (x,y)\neq(0,0)\ &0 \qquad ;\text{ for } (x,y)=(0,0)\ \end{cases}$$ – Rag Aug 29 '17 at 07:16