3.14. Find such functions $\xi(x), \eta(x), \omega(y), \tau(y)$, so that the function of a complex variable

$$
f(z)=\left(\xi(x)+\omega(y)-3 x y^{2}\right)+i\left(\eta(x)+\tau(y)+3 x^{2} y+4 x y+5 x\right)
$$

is analytic. Find its form.
Solution. Since the function $f(z)$ must be analytic, it must satisfy the Cauchy-Riemann conditions, i.e.,

$$
\left\{\begin{array}{l}
\frac{\partial}{\partial x}\left(\xi(x)+\omega(y)-3 x y^{2}\right)=\frac{\partial}{\partial y}\left(\eta(x)+\tau(y)+3 x^{2} y+4 x y+5 x\right) \\
\frac{\partial}{\partial y}\left(\xi(x)+\omega(y)-3 x y^{2}\right)=-\frac{\partial}{\partial x}\left(\eta(x)+\tau(y)+3 x^{2} y+4 x y+5 x\right)
\end{array}\right.
$$

or, equivalently:

$$
\left\{\begin{array}{l}
\frac{d \xi(x)}{d x}-3 y^{2}=\frac{d \tau(y)}{d y}+3 x^{2}+4 x \\
\frac{d \omega(y)}{d y}=-\frac{d \eta(x)}{d x}-4 y-5
\end{array}\right.
$$

From this, we have:

$$
\begin{gathered}
\left\{\begin{array}{l}
\frac{d \xi(x)}{d x}-3 x^{2}-4 x=\frac{d \tau(y)}{d y}+3 y^{2}, \\
\frac{d \eta(x)}{d x}=-4 y-5-\frac{d \omega(y)}{d y} ;
\end{array} \Leftrightarrow\right. \\
\Leftrightarrow\left\{\begin{array}{l}
\int_{a}^{x}\left(\frac{d \xi(t)}{d t}-3 t^{2}-4 t\right) d t=\int_{b}^{y}\left(\frac{d \tau(t)}{d t}+3 t^{2}\right) d t, \\
\int_{c}^{x} \frac{d \eta(t)}{d t} d t=\int_{d}^{y}\left(-4 t-5-\frac{d \omega(t)}{d t}\right) d t ;
\end{array} \Leftrightarrow\right. \\
\Leftrightarrow\left\{\begin{array}{c}
\xi(x)-x^{3}-2 x^{2}-\xi(a)+a^{3}+2 a^{2}=\tau(y)+y^{3}-\tau(b)-b^{3}, \\
\eta(x)-\eta(c)=-2 y^{2}-5 y-\omega(y)+2 d^{2}+5 d+\omega(d) ;
\end{array} \Leftrightarrow\right. \\
\Leftrightarrow\left\{\begin{array}{c}
\xi(x)-x^{3}-2 x^{2}+C_{1}=\tau(y)+y^{3},(1) \\
\eta(x)+C_{2}=-2 y^{2}-5 y-\omega(y) .
\end{array}\right.
\end{gathered}
$$

Equality (1) is possible if

$$
\left\{\begin{array}{l}
\xi(x)=x^{3}+2 x^{2}+C_{3}, \\
\tau(y)=-y^{3}+C_{4} .
\end{array}\right.
$$

Equality (2) is possible if

$$
\left\{\begin{array}{l}
\eta(x)=C_{5} \\
\omega(y)=-2 y^{2}-5 y+C_{6}
\end{array}\right.
$$

Thus, we have:

$$
\begin{gathered}
f(z)=\left(x^{3}+2 x^{2}-2 y^{2}-5 y-3 x y^{2}\right)+ \\
+i\left(-y^{3}+3 x^{2} y+4 x y+5 x\right)+z_{0}, z_{0}=\sigma_{r}+i \sigma_{i} \in \approx
\end{gathered}
$$

Suppose the function $f(z)$ has the form: $f(z)=\alpha z^{3}+\beta z^{2}+\gamma z+\delta$, where the coefficients $\alpha, \beta, \gamma, \delta \in \mathbb{\sim}$ need to be found. We separate $\operatorname{Re}\left(\alpha z^{3}+\beta z^{2}+\gamma z+\delta\right)$ and $\operatorname{Im}\left(\alpha z^{3}+\beta z^{2}+\gamma z+\delta\right)$, setting $Q=$ $=Q_{r}+i Q_{i}, Q=\alpha, \beta, \gamma, \delta, Q_{r}, Q_{i} \in \mathbb{R}$, we have:

$$
\begin{gathered}
\quad f(z)=\alpha z^{3}+\beta z^{2}+\gamma z+\delta=\left(\alpha_{r}+i \alpha_{i}\right)(x+i y)^{3}+ \\
+\left(\beta_{r}+i \beta_{i}\right)(x+i y)^{2}+\left(\gamma_{r}+i \gamma_{i}\right)(x+i y)+\left(\delta_{r}+i \delta_{i}\right)= \\
=\left\{\alpha_{i} y^{3}-3 \alpha_{r} x y^{2}-3 \alpha_{i} x^{2} y-\beta_{r} y^{2}-2 \beta_{i} x y-\gamma_{i} y+\alpha_{r} x^{3}+\right. \\
\left.+\beta_{r} x^{2}+\gamma_{r} x+\delta_{r}\right\}+i\left\{-\alpha_{r} y^{3}-3 \alpha_{i} x y^{2}-\beta_{i} y^{2}+3 \alpha_{r} x^{2} y+\right. \\
\left.\quad+\alpha_{i} x^{3}+2 \beta_{r} x y+\beta_{i} x^{2}+\gamma_{r} y+\gamma_{i} x+\delta_{i}\right\} .
\end{gathered}
$$

From the principle of equality of two complex numbers, we have:

$$
\left\{\begin{array}{l}
x^{3}+2 x^{2}-2 y^{2}-5 y-3 x y^{2}+\sigma_{r}=\alpha_{i} y^{3}-3 \alpha_{r} x y^{2}-3 \alpha_{i} x^{2} y- \\
\quad-\beta_{r} y^{2}-2 \beta_{i} x y-\gamma_{i} y+\alpha_{r} x^{3}+\beta_{r} x^{2}+\gamma_{r} x+\delta_{r} \\
-y^{3}+3 x^{2} y+4 x y+5 x+\sigma_{i}=-\alpha_{r} y^{3}-3 \alpha_{i} x y^{2}-\beta_{i} y^{2}+ \\
\quad+3 \alpha_{r} x^{2} y+\alpha_{i} x^{3}+2 \beta_{r} x y+\beta_{i} x^{2}+\gamma_{r} y+\gamma_{i} x+\delta_{i}
\end{array}\right.
$$

From this, we conclude that:

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-210.jpg?height=213&width=555&top_left_y=1276&top_left_x=376)

Thus, we have: $\alpha=1+i 0, \beta=2+i 0, \gamma=0+i 5$, $z_{0}=\sigma_{r}+i \sigma_{i}, \sigma_{r}, \sigma_{i} \in \mathbb{R}$, which means, $f(z)=z^{3}+2 z^{2}+5 i z+z_{0}$.