Find the line integral along curve $C$ of $[f(z)]^2=z$ where $f(1)=1$.
Here is curve c: https://i.stack.imgur.com/fNrIA.jpg (Sorry for the blur, the points are $1$ and $e$)
How can I solve this? I am lost. Is $f(z)= z^{1/2}$? If so, is $f(1) = 1$ telling us the radius in the beginning is $1$? Which method do we use to solve this and how can you tell? Note: No residue method.
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I'm interpreting the problem as follows: In the neighborhood of the starting point of $\gamma$ we should assume $f(z):={\rm pv}\sqrt{z}$. Proceeding along $\gamma$ the function $f$ is understood as analytic continuation of this original "germ". The curve $\gamma$ may safely be replaced by the concatenation of $$\gamma_1:\quad t\mapsto e^{it}\qquad(0\leq t\leq 4\pi)$$ and $$\gamma_2:\quad x\mapsto x\qquad(1\leq x\leq e)\ .$$ Along $\gamma_1$ we obviously have $$f\bigl(z(t)\bigr)=e^{it/2}\ .$$ After two full turns around $0$ the the analytic continuation of $f$ has arrived again at $f(z):={\rm pv}\sqrt{z}$. It follows that along $\gamma_2$ we simply have $$f\bigl(z(x)\bigr)=\sqrt{x}\ .$$ In this way we obtain $$\int_\gamma f(z)\>dz=\int_0^{4\pi} e^{it/2}\>ie^{it}\>dt+\int_1^e \sqrt{x}\>dx={2\over3}(e^{2/3}-1)\ .\tag{1}$$ It may be of interest to consider also the analogous curve $\gamma'$ winding three times around $0$ before drifting off to $e$. At the end of $\gamma_1'$ the analytic continuation of $f$ arrives at $f(z)=-{\rm pv}\sqrt{z}$, and instead of $(1)$ we obtain $$\int_{\gamma'}f(z)\>dz=\int_0^{6\pi} e^{it/2}\>ie^{it}\>dt+\int_1^e(-\sqrt{x})\>dx=-{4\over3}-{2\over3}(e^{2/3}-1)\ .$$
$f(z)$ is one of the square roots of $z$. Since $f(1)=1$, we deduce that $f(z)=\sqrt z$. Since $f$ is continuous and has a primitive in the curve $C$, which is $F(z)=\frac{2}{3}z^{\frac{3}{2}}$, you can calculate the integral simply by:
$$\int_{C}f(z)dz=F(e)-F(1)=\frac{2}{3}(e^{\frac{2}{3}}-1)$$