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I want to show that the function $f(z) = \sqrt{z}$ is analytic on $D = \{z\in\mathbb{C}:Re(z)>0\}$ by the Cauchy-Riemann equation.

But here is the thing. I fail to rewrite $f(z)$ into $u(x,y) + iv(x,y)$. So can anyone help me to rewrite it?

Is $\sqrt{z}$ an analytic function? Here I found that $f(z)$ is analytic, but they did not use Cauchy-Riemann which I want to use.

Thanks in advance.

Math is like Friday
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  • I can write $f(z) = r^{1/2}\cdot e^{I\cdot \frac{\theta}{2}}$, so $f(z) = \sqrt(r)\cdot cos(\frac{\theta}{2}) + \sqrt{r} i\cdot sin(\frac{\theta}{2})$. And again I stuck here – Math is like Friday Apr 06 '20 at 12:33
  • You can't use Cauchy-Riemann for functions $f \colon \mathbb R \to \mathbb R$. The derivative $\partial f / \partial y$ will always be zero while $\partial f / \partial x$ won't. – Targon Apr 06 '20 at 12:38
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    @Targon I don't quite understand your comment. He is considering the square root of a complex number. He tries to rewrite it as function of the real and the imaginary part. This is exactly what we need to do to be able to apply the CR-equations. No? – Severin Schraven Apr 06 '20 at 12:40
  • Note that $r=\sqrt{x^2+y^2}$ and $\cos(\theta)= y/\sqrt{x^2+y^2}$. Similarly we can express $\sin(\theta)$. – Severin Schraven Apr 06 '20 at 12:42

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I think $\sqrt{(x+\sqrt{x^2+y^2})/2}+i\,\text{sgn}(y)\sqrt{(-x+\sqrt{x^2+y^2})/2}$ is the expression for the 'positive' root of $x+iy$ that you are looking for. See e.g. here for more info. I'm not sure that it'll be economical to prove the holomorphicity of $\sqrt{z}$ by verifying the Cauchy-Riemann equations starting from this expression.

5th decile
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