Example 2. Calculate the integral

$$
\int_{L}|z| \bar{z} d z
$$

where $L-$ is the upper semicircle $|z|=1, \operatorname{Re} z>0$ traversed counterclockwise.
Solution. In this case, it is convenient to use the parametric form of the curve $L$ as $z=e^{i t}(0 \leqslant t \leqslant \pi)$ and apply formula (2)

$$
\int_{L} f(z) d z=\int_{\alpha}^{\beta} f(z(t)) z^{\prime}(t) d t
$$

1. Find $\bar{z}=e^{-i t},|z|=1, d z=i e^{i t} d t$.
2. Substitute into the integrand and compute the integral

$$
\int_{L}|z| \bar{z} d z=\int_{0}^{\pi} e^{-i t} i e^{i t} d t=\int_{0}^{\pi} i d t=i \pi
$$

Answer. $\int_{L}|z| \bar{z} d z=i \pi$.

Conditions of the Problems. Compute the integrals of functions of a complex variable along the given curves.

1. $\int_{L} \bar{z} d z, L$ - the semicircle $|z|=1, \operatorname{Im} z \geqslant 0$.
2. $\int_{L} z \operatorname{Re} z d z, L-$ the line segment from point $z_{1}=0$ to point $z_{2}=i$.
3. $\int_{L} z^{2}|z| d z, L$ - the semicircle $|z|=2, \operatorname{Im} z \leqslant 0$.
4. $\int_{L} z \operatorname{Im} z d z, L-$ the line segment from point $z_{1}=0$ to point $z_{2}=1+i$.
5. $\int_{L} e^{|z|} d z, L-$ the semicircle $|z|=1, \operatorname{Re} z \leqslant 0$.
6. $\int_{L} \operatorname{Im} z d z, L-$ the semicircle $|z|=1, \operatorname{Im} z \geqslant 0$.
7. $\int_{L} \bar{z} d z, L-$ the line segment from point $z_{1}=0$ to point $z_{2}=1+i$.
8. $\int_{L}|z| \operatorname{Re} z d z, L-$ the semicircle $|z|=3, \operatorname{Im} z \leqslant 0$.
9. $\int_{L} z \sin |z| d z, L-$ the semicircle $|z|=2, \operatorname{Re} z \geqslant 0$.
10. $\int_{L} z^{2} \operatorname{Im} z d z, L-$ the line segment from point $z_{1}=0$ to point $z_{2}=2+2 i$.

Answers. 1. $\pi i$. 2. $0.3 .3 .-16 \pi / 3$. 4. $2 i / 3$. 5. $-2 e i$. $6 .-\pi / 2$. 7. 1. 8. $-27 \pi i / 2$. 9. 0. $10.8+8 i$.

## 1.6. Integral of an Analytic Function

Problem Statement. Compute the integral

$$
\int_{L} f(z) d z
$$

where $f(z)$ is a function analytic in a simply connected domain $D$, and $L$ is a piecewise smooth curve entirely lying in the domain $D$ and connecting points $z_{1}$ and $z_{2}$.

Plan of Solution. The integral along a piecewise smooth curve $L \in D$ of a function analytic in a simply connected domain $D$ does not depend on the path of integration but only on the initial and final points $z_{1}$ and $z_{2}$ and can be computed using the Newton-Leibniz formula

$$
\int_{L} f(z) d z=\int_{z_{1}}^{z_{2}} f(z) d z=F\left(z_{2}\right)-F\left(z_{1}\right)
$$

where $F(z)$ is the antiderivative of the function $f(z)$.

1. Find the antiderivative $F(z)$ using table integrals, properties of integrals, and methods known from mathematical analysis.
2. Compute the integral using the Newton-Leibniz formula (1). Write the answer in algebraic form.