3.17. Calculate

$$
\int_{L}\left(2 x^{2}+3 y^{2} i\right) d z
$$

where $L-$ is the straight line segment connecting the points $z_{1}=1+i, z_{2}=2+3 i$.
Solution. Let's find the equation of the line passing through the points $z_{1}=1+i$ and $z_{2}=2+3 i$. Assume it has the form $\operatorname{Im} z=k \operatorname{Re} z+b$, from which it follows that the system is valid:

$$
\left\{\begin{array} { l } 
{ \operatorname { I m } z _ { 1 } = k \operatorname { R e } z _ { 1 } + b , } \\
{ \operatorname { I m } z _ { 2 } = k \operatorname { R e } z _ { 2 } + b ; }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 1 = k + b , } \\
{ 3 = 2 k + b }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
k=2 \\
b=-1
\end{array}\right.\right.\right.
$$

Thus, the desired line is $\operatorname{Im} z=2 \operatorname{Re} z-1$, then the contour $L$ can be represented as follows:

$$
L=\{z \in \mathbb{\mathbb { S }} \mid z=x+i y, x=t, y=2 t-1, t \in[1,2]\}
$$

Now we can easily compute the integral

$$
\begin{gathered}
\int_{L}\left(2 x^{2}+3 y^{2} i\right) d z: \\
\iint_{L}\left(2 x^{2}+3 y^{2} i\right) d z=\left\{\begin{array}{l}
x=t, y=2 t-1 \\
d z=d x+i d y=(1+2 i) d t \\
2 x^{2}+3 y^{2} i=2 t^{2}+3(2 t-1)^{2} i
\end{array}\right\}= \\
=\int_{1}^{2}\left(2 t^{2}+3(2 t-1)^{2} i\right)(1+2 i) d t= \\
=\left.(1+2 i)\left(\frac{2}{3} t^{3}+\frac{1}{2}(2 t-1)^{3} i\right)\right|_{1} ^{2}=\frac{1}{3}(-64+67 i)
\end{gathered}
$$