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First of all, is $z\mapsto \sqrt{z}$ really a function? Indeed, each complex number has two square roots, so I have the impression that $\sqrt z$ is not well-defined as a function, is it? Then, I was wondering if $\sqrt z=z^{\frac{1}{2}}$ was still true. Indeed, $z^{1/2}:=e^{\frac{1}{2}\log(z)}$, and this is holomorphic in some domain. Whereas, I even have doubt if $z\mapsto \sqrt z$ is really a function. What do you think ?

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The notations $\sqrt{z}$ and $z^{1/2}$ both refer to the principal square root of $z$, which is defined as the solution of $w^2=z$ with argument in $(-\pi/2,\pi/2]$. Using $\log(z)$ you run into the same problem because $\exp(w)=z$ even has infinitely many solutions $w\in\Bbb C$. This is why for complex numbers you would write $\operatorname{Log}(z)$ to refer to the principal value of the logarithm, which is the solution of $\exp(w)=z$ with imaginary part in $(-\pi,\pi]$.

Note that the two choices of principal branches agree so that indeed $$ \sqrt{z} = \exp\left(\frac12 \operatorname{Log}(z)\right) = z^{1/2}. $$

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