Theorem 1   Let the complex function $f$ be analytic in a simply connected open domain $U$ of the complex plane, and let $a$ and $b$ be any two points of $U$ . Then the contour integral

(1)

is independent on the path $\gamma$ which in $U$ goes from $a$ to $b$ .

Example. Let's consider the integral (1) of the real part function defined by $$f(z) := \mbox{Re}(z)$$ with the path $\gamma$ going from the point $O = (0,\,0)$ to the point $Q = (1,\,1)$ . If $\gamma$ is the line segment $OQ$ , we may use the substitution $$z := (1\!+\!i)t,\quad dz = (1\!+\!i)\,dt,\quad 0 \leqq t \leqq 1,$$ and (1) equals $$\int_0^1t\!\cdot\!(1\!+\!i)\,dt = \frac{1}{2}\!+\!\frac{1}{2}i.$$ Secondly, we choose for $\gamma$ the broken line $OPQ$ where $P = (1,\,0)$ . Now (1) is the sum $$\int_{OP}\mbox{Re}(z)\,dz+\int_{PQ}\mbox{Re}(z)\,dz = \int_0^1x\,dx+\int_0^1i\,dy = \frac{1}{2}\!+\!i.$$ Thus, the integral (1) of the function depends on the path between the two points. This is explained by the fact that the real part function $f$ is not analytic -- its real part $x$ and imaginary part 0 do not satisfy the Cauchy-Riemann equations.