Example 8. The point $z=x+i y$ describes the segment

$$
x=1, \quad-1 \leqslant y \leqslant 1
$$

What is the length of the line obtained by mapping this segment using the function $w=z^{2}$?
Solution. First method. We have $w=z^{2}$, or

$$
u+i v=x^{2}-y^{2}+i 2 x y
$$

i.e.

$$
\left\{\begin{array}{l}
u=x^{2}-y^{2} \\
v=2 x y
\end{array}\right.
$$

Obviously, on the line (10) we will have

$$
\left\{\begin{array}{l}
u=1-y^{2} \\
v=2 y
\end{array}\right.
$$

where, as $y$ changes from -1 to +1, $v$ will change from -2 to +2. From (11) we obtain the equation of the parabola (Fig. 3)

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-036.jpg?height=757&width=333&top_left_y=850&top_left_x=881)

Fig. 3

$$
u=1-\frac{v^{2}}{4}
$$

The length of the arc $\boldsymbol{A}^{\prime} \boldsymbol{B}^{\prime} \boldsymbol{C}^{\prime}$ of the parabola (12)

$$
t_{w}=2 \int_{0}^{2} \sqrt{1+\frac{v^{2}}{4}} d v=\int_{0}^{2} \sqrt{4+v^{2}} d v=2 \sqrt{2}+\ln (3+2 \sqrt{2})
$$

Second method. Using formula (8), we will have

$$
\begin{aligned}
l_{w} & =\int_{L}\left|f^{\prime}(z)\right| \cdot|d z|=\int_{L}|2 z| \cdot|d z|= \\
& =2 \int_{-1}^{1} \sqrt{1+y^{2}} d y=4 \int_{0}^{1} \sqrt{1+y^{2}} d y=2 \sqrt{2}+\ln (3+2 \sqrt{2})
\end{aligned}
$$