Help, why these are two different results of integral of $\sqrt{z}$ on unit circle depending the choice of Branch cut everyone, I want test the effect of different choice of branch cut for contour, So I find a simple function, i.e. $\sqrt{z}$ with $z=re^{i\theta}$ on 1st Branch as $$I=\oint_{UnitCircle}{\sqrt{z}dz}$$ 
I always find two contours to calculate and contrast.
The 1st one contour is shown as 
By using the Cauchy Theorem, I have $\oint_{\Gamma}{\sqrt{z}dz}=\int_{C_1}+\int_{C_2}+\int_{C_3}+\int_{C}$, The $\int_{C}$ is what I want when the radius $\delta$ of small circle turn to $0$. And I have three results for other three contours which are given as
$$\int_{C_{1}}{\sqrt{z}dz}=\int_{C_{1}}{\sqrt{r}e^{i\frac{0}{2}}d(re^{i0})}=\int_{C_{1}}{\sqrt{r}dr}=\int_{0}^{1}{\sqrt{r}dr}=\frac{2}{3}$$
$$\int_{C_{2}}{\sqrt{z}dz}=\int_{C_{2}}{\sqrt{r}e^{i\frac{2\pi}{2}}d(re^{i 2\pi)}}=-\int_{C_{2}}{\sqrt{r}dr}=-\int_{1}^{0}{\sqrt{r}dr}=\frac{2}{3}$$
For $C_{3}$, I use the $\epsilon-\Delta$ formula as: given positive real number $\epsilon$, have a positive real number $\Delta<\epsilon^{\frac{2}{3}}$, when $\left|z-0\right|=\left|z\right|=\delta$, $\left|zf(z)-0\right|=\left|r^{\frac{3}{2}}e^{\frac{i3\theta}{2}}\right|=r^{\frac{3}{2}}<\Delta^{\frac{3}{2}}<\epsilon$, so I have
$$lim_{\delta\rightarrow 0}{\int_{C_{3}}\sqrt{z}dz}=i0(2\pi-0)=0$$.
So, I directly get $$I=-\frac{4}{3}$$.
Then, I re-calculate $I$ using 2nd contour as shown as , and I get a different result. I also start with the Cauchy Theorem as $\oint_{\Gamma}{\sqrt{z}dz}=\int_{C_1}+\int_{C_2}+\int_{C_3}+\int_{C}$, The $\int_{C}$ is what I want when the radius $\delta$ of small circle turn to $0$. And I have three results for other three contours which are given as
$$\int_{C_{1}}{\sqrt{z}dz}=\int_{C_{1}}{\sqrt{r}e^{i\frac{\pi}{2}}d(re^{i\pi})}=\int_{C_{1}}{-i\sqrt{r}dr}=-i\int_{1}^{0}{\sqrt{r}dr}=i\int_{0}^{1}{\sqrt{r}dr}=i\frac{2}{3}$$
$$\int_{C_{2}}{\sqrt{z}dz}=\int_{C_{2}}{\sqrt{r}e^{i\frac{-\pi}{2}}d(re^{-i\pi})}=\int_{C_{2}}{-i\sqrt{r}d(-r)}=i\int_{C_{2}}{\sqrt{r}dr}=i\int_{0}^{1}{\sqrt{r}dr}=i\frac{2}{3}$$
And I also have $\epsilon-\Delta$ formula:given positive real number $\epsilon$, have a positive real number $\Delta<\epsilon^{\frac{2}{3}}$, when $\left|z-0\right|=\left|z\right|=\delta$, $\left|zf(z)-0\right|=\left|r^{\frac{3}{2}}e^{\frac{i3\theta}{2}}\right|=r^{\frac{3}{2}}<\Delta^{\frac{3}{2}}<\epsilon$, which gives
$$lim_{\delta\rightarrow 0}{\int_{C_{3}}\sqrt{z}dz}=i0(\pi+\pi)=0$$.
So, I get $$I=-i\frac{4}{3}$$ which is not same as the result given in 1st contour, Is it correct? 
I know that the complex contour integral depends the kink and path at same time, But the result I wanted is calculate on a "Closed" circle in a roughly view. On the other hand, the circle in these two contours actually start/end at different point, i.e. $(0, 2\pi]$ and $(-\pi, \pi]$. May be it is the key point. Can anyone help me fix it? Thanks!!Please!!
 A: The square root isn't single-valued in any neighborhood of the origin. When you make a slit (or "branch cut"), you can choose a single-valued holomorphic function on the complement of the slit, but:


*

*The values of that function at non-zero $z$ depend on the angle of the ray where you make the cut;

*No choices can agree everywhere off the two slits.
The diagram shows the real part of a branch of square root with the positive real axis removed (left) and the real part of a branch of square root with the negative real axis removed (right). The two branches agree in the lower half-plane (toward the left), but differ by a sign in the upper half-plane (toward the right).

The underlying analytic issue comes from choosing a branch of polar angle: If $\theta$ is a branch of angle on a slit plane, then
$$
f(re^{i\theta}) = \sqrt{r} e^{i\theta/2}
$$
is the corresponding branch of square root.


*

*If you remove the positive axis, there are (essentially) two choices: $0 < \theta < 2\pi$, or $2\pi < \theta < 4\pi$.

*If you remove the negative axis, there are again two choices: $-\pi < \theta < \pi$, or $\pi < \theta < 3\pi$.
It's easy to check that no matter how you choose branches of $\theta$, you can't get a single ("compatible choice of") square root off the real axis.
The relevance for the phenomenon you've noticed is: Your integral is really measuring the "jump" of the chosen branch on opposite sides of the slit.
Geometrically, "integrating the square root around the unit circle" isn't a well-defined operation: The unit circle does not determine a closed curve on the Riemann surface of the square root, and the value of the integral depends on where your path starts.
