contour integrating 1/z over unit square equal zero?

Question

After I have studied contour integration and related theorems for some time, I encounter a quite confusing problem:

integrating $\frac{1}{z}$ about a unit square centered at the origin(positive orientation).

Let $F(z)$ be the antiderivative of $\frac{1}{z}$, seperating the contour into 4 straight lines, then

$$\int_\Gamma \frac{1}{z}dz = (F(0.5+0.5i) - F(0.5-0.5i)) + (F(-0.5+0.5i) - F(0.5+0.5i)) +... = 0$$

I expect the answer to be $2\pi i$ however that results in zero. What mistake did I make?

p.s. please do not use residue theorem or any theorems to directly show that the integral equal $2\pi i$. I would like to see some sort of direct integration. Thanks.

Answer 1

Observe the domain (i mean simply connected region) where the branch of logarithm fn is analytic.Now see if you take closed curve which is homotopic to a circle in that simply connected region.And then you do integral and you find it will be 2 *pi *i *n where n is a positive integer,it is the winding number,it depends no. of times of rotation on the circle or the closed curve.

Answer 2

Hint: $F(z)=log(z)$ has a branch cut. See the Wikipedia page on complex logarithm.