# Give principal branch of complex function

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 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. – copper.hat Feb 3 at 4:01

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 $$1-2=-1$$.