3.13. Given the imaginary part $v(x, y)=3^{x} \sin (y \ln 3)$ of the analytic function $f(z)$. Find $f(z)$.
Solution.

Since the function $f(z)$ is analytic, the Cauchy-Riemann conditions are satisfied, i.e.,

$$
\frac{\partial u(x, y)}{\partial x}=\frac{\partial v(x, y)}{\partial y}, \quad \frac{\partial u(x, y)}{\partial y}=-\frac{\partial v(x, y)}{\partial x}
$$

For our problem, these conditions can be rewritten as:

$$
\frac{\partial u(x, y)}{\partial x}=3^{x} \ln 3 \cos (y \ln 3), \quad \frac{\partial u(x, y)}{\partial y}=-3^{x} \ln 3 \sin (y \ln 3)
$$

The expression

$$
\frac{\partial u(x, y)}{\partial x} d x+\frac{\partial u(x, y)}{\partial y} d y
$$

represents the total differential of some function $U(x, y)$. Indeed, the necessary condition for this,

$$
\frac{\partial}{\partial y} \frac{\partial u(x, y)}{\partial x}=\frac{\partial}{\partial x} \frac{\partial u(x, y)}{\partial y}
$$

holds in this case. We will reconstruct the potential function $U(x, y)$ from its total differential by taking a line integral of the second kind over the contour $L=L_{1}+L_{2}$, where

$$
L_{1}:\left\{\begin{array}{l}
x=t, \\
y=y_{0}, \\
t \in\left[x_{0}, x\right] ;
\end{array} \quad L_{2}:\left\{\begin{array}{l}
x=x \\
y=t \\
t \in\left[y_{0}, y\right]
\end{array}\right.\right.
$$

We have:

$$
\begin{gathered}
U(x, y)=\int_{L} \frac{\partial u(x, y)}{\partial x} d x+\frac{\partial u(x, y)}{\partial y} d y= \\
=\int_{L} 3^{x} \ln 3 \cos (y \ln 3) d x-3^{x} \ln 3 \sin (y \ln 3) d y= \\
=\int_{L_{1}} 3^{t} \ln 3 \cos \left(y_{0} \ln 3\right) d t-\int_{L_{2}} 3^{x} \cdot \ln 3 \sin (t \ln 3) d t= \\
=\cos \left(y_{0} \ln 3\right) \int_{L_{1}} 3^{t} \ln 3 d t-3^{x} \int_{L_{2}} \ln 3 \sin (t \ln 3) d t= \\
=\cos \left(y_{0} \ln 3\right) \int_{x_{0}}^{x} d 3^{t}+3^{x} \int_{y_{0}}^{y} d \cos (t \ln 3)=\cos \left(y_{0} \ln 3\right) 3^{x}- \\
-\cos \left(y_{0} \ln 3\right) 3^{x_{0}}+3^{x} \cos (y \ln 3)-3^{x} \cos \left(y_{0} \ln 3\right)= \\
=3^{x} \cos (y \ln 3)-3^{x_{0}} \cos \left(y_{0} \ln 3\right)=U_{1}(x, y)-U_{1}\left(x_{0}, y_{0}\right) .
\end{gathered}
$$

From this, it is clear that

$$
U(x, y)=3^{x} \cos (y \ln 3)+C, \quad C \in \mathbb{R}
$$

It is not difficult to see that this is the real part of the function $f(z)$, from which it follows that:

$$
\begin{gathered}
f(z)=3^{x} \cos (y \ln 3)+C+i 3^{x} \sin (y \ln 3)= \\
=3^{x}(\cos (y \ln 3)+i \sin (y \ln 3))+C= \\
=3^{x} e^{i y \ln 3}+C=3^{x} 3^{i y}+C=3^{x+i y}+C=3^{z}+C .
\end{gathered}
$$