Example 1. Compute the integral

$$
\int_{C}(1+i-2 \bar{z}) d z
$$

along the lines connecting the points $z_{1}=0$ and $z_{2}=1+i$,

1) along a straight line;
2) along the parabola $y=x^{2}$;
3) along the broken line $z_{\mathrm{j}} z_{3} z_{2}$, where $z_{3}=1$.
Solution. Rewrite the integrand function as

$$
1+i-2 \bar{z}=(1-2 x)+i(1+2 y)
$$

Here $u=1-2 x, v=1+2 y$.

Applying formula (1), we get

$$
\int_{C}(1+i-2 \bar{z}) d z=\int_{C}(1-2 x) d x-(1+2 y) d y+i \int_{C}(1+2 y) d x+(1-2 x) d y .
$$

1) The equation of the line passing through the points $z=0$ and $z_{2}=1+i$ is $y=x, 0 \leqslant x \leqslant 1$, so $d y=d x$. Therefore,

$$
\begin{array}{r}
\int_{C}(1+i-2 \bar{z}) d z=\int_{0}^{1}[(1-2 x)-(1+2 x)] d x+ \\
+i \int_{0}^{1}[(1+2 x)+(1-2 x)] d x=2(i-1)
\end{array}
$$

2) For the parabola $y=x^{2}$, we have $d y=2 x d x(0 \leqslant x \leqslant 1)$. Therefore,

$$
\begin{aligned}
& \int_{C}(1+i-2 \bar{z}) d z=\int_{0}^{1}\left[1-2 x-\left(1+2 x^{2}\right) 2 x\right] d x+ \\
& \quad+i \int_{0}^{1}\left[1+2 x^{2}+(1-2 x) 2 x\right] d x=-2+\frac{4}{3} i
\end{aligned}
$$

3) On the segment $z_{1} z_{3}: y=0, d y=0,0 \leqslant x \leqslant 1$. On the segment $z_{3} z_{2}: x=1$, $d x=0,0 \leqslant y \leqslant 1$. Using the linearity property of line integrals, we get

$$
\begin{aligned}
\int_{C}(1+i-2 \bar{z}) d z & =\int_{z_{1} z_{3}}(1+i-2 \bar{z}) d z+\int_{z_{3} z_{2}}(1+i-2 \bar{z}) d z= \\
& =\int_{0}^{1}(1-2 x) d x+i \int_{0}^{1} d x-\int_{0}^{1}(1+2 y) d y+i \int_{0}^{1}(1-2 \cdot 1) d y=-2 .
\end{aligned}
$$

This example shows that the integral of a continuous but non-analytic function generally depends on the path of integration.