Example 21. Using the Joukowsky function, find the image of the region

$$
0<|z|<1, \quad 0<\arg z<\frac{\pi}{4}
$$
Solution. Substitute $z=r e^{i \varphi}$ into the Joukowski function

$$
w=\frac{1}{2}\left(z+\frac{1}{z}\right)
$$

and separate the real and imaginary parts; we get

$$
\left\{\begin{array}{l}
u=\frac{1}{2}\left(r+\frac{1}{r}\right) \cos \varphi \\
v=\frac{1}{2}\left(r-\frac{1}{r}\right) \sin \varphi
\end{array}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-140.jpg?height=400&width=543&top_left_y=968&top_left_x=685)

Traversing the contour of the sector in the positive direction $O A B O$ (Fig. 26), we get: the segment $O A$ will transform into an infinite segment of the real axis, traversed from $u=+\infty$ to $u=1$ (Fig. 27); the arc $A B$ will transform into the segment of the real axis $A^{\prime} B^{\prime}$, and the segment $B O$ will transform into the curve

$$
\begin{aligned}
& u=\frac{\sqrt{2}}{4}\left(r+\frac{1}{r}\right) \\
& v=\frac{\sqrt{2}}{4}\left(r-\frac{1}{r}\right)
\end{aligned}
$$

or $u^{2}-v^{2}=\frac{1}{2}$ (a hyperbola).

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-140.jpg?height=41&width=494&top_left_y=1606&top_left_x=104)
![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-140.jpg?height=480&width=602&top_left_y=1426&top_left_x=636)

According to the principle of one-to-one correspondence of boundaries, we get that the given sector is transformed by the Joukowski function into the region

$$
u^{2}-v^{2}>\frac{1}{2}, \quad u>\frac{\sqrt{2}}{2}, \quad v<0
$$

$\nabla$