# In $\mathbb C$ does $\sqrt z=z^{1/2}$?

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 ?

## 1 Answer

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 $$\sqrt{z} = \exp\left(\frac12 \operatorname{Log}(z)\right) = z^{1/2}.$$