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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.

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  • $\begingroup$ The principal branch is $z \mapsto e^{{1 \over 2} Log (1-z)}$, defined on $\mathbb{C} \setminus [1,\infty)$. Note the change in the real axis part of the definition. $\endgroup$ – copper.hat Feb 3 at 4:01
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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 $1-2=-1$.

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