Give the principal branch of complex-valued function $f(z)=\sqrt{1-z}$.
My approach: We know that square root is multi-valued function.
Consider the set $D:=\mathbb{C}-\{z:\Im(z)=0, \Re(z)\leq 1\}$ which is region (open and connected subset) in $\mathbb{C}$. Let's take the principal branch of logarithm, i.e. $$\text{Log}(1-z)=\log|1-z|+i\arg(1-z).$$ Then the principal branch of $f(z)$ will be $$f_1(z)=\exp\left(\frac{1}{2}(\log|1-z|+i\arg(1-z))\right)$$ on the region $D$.
Also $(f_1(z))^2=\exp\left(\log|1-z|+i\arg(1-z)\right)=\exp(\text{Log}(1-z))=1-z$.
We see that function $f_1(z)$ is single-valued on $D$, continuous on $D$ and also it's one of the values of function $\sqrt{1-z}$.
Please, can anyone take a look at my solution and say is it ok? This the first time when I have solved problems on complex analysis. I've solved it after many hours of thinking and reading lecture materials. So please do not duplicate my question.
The principle branch of logarithm is usually defined on $\mathbb C \setminus \{z:Im(z)=0, Re (z) \leq 0\}$. If this is what you mean by Log then the correct answer is $\mathbb C \setminus \{z:Im(z)=0, Re (z) \geq 1\}$. Note that $z=2$ is in the region you are considering but the principle branch of log is not defined at $1-2=-1$.