second form of Cauchy integral theorem

Theorem.

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

${\displaystyle {\int }_{\gamma }}f(z)𝑑z$(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

$$f(z):=\text{Re}(z)$$

with the path $\gamma$ going from the point  $O=(0,\mathrm{\hspace{0.17em}0})$  to the point  $Q=(1,\mathrm{\hspace{0.17em}1})$.  If $\gamma$ is the line segment $OQ$, we may use the substitution

$$z:=(1+i)t,dz=(1+i)dt,0\leqq t\leqq 1,$$

and (1) equals

$${\int }_0^{1}t\cdot (1+i)𝑑t=\frac{1}{2}+\frac{1}{2}i.$$

Secondly, we choose for $\gamma$ the broken line $OPQ$ where  $P=(1,\mathrm{\hspace{0.17em}0})$.  Now (1) is the sum

$${\int }_{OP}\text{Re}(z)𝑑z+{\int }_{PQ}\text{Re}(z)𝑑z={\int }_0^{1}x𝑑x+{\int }_0^{1}i𝑑y=\frac{1}{2}+i.$$

Thus, the integral (1) of the function depends on the path between the two points.  This is explained by the fact that the function $f$ is not analytic — its real part $x$ and imaginary part 0 do not satisfy the Cauchy-Riemann equations.