3.6. Find the mapping of the circle $x^{2}+y^{2}=4$ onto the Ouv plane using the function $w=3z+2$.
Solution. The function $w=3z+2$ maps circles centered at $z=0$ to circles centered at $w=0$; lines passing through the point $z=0$ to lines passing through the point $w=0$, indeed:

$$
w=3z+2=3\rho e^{i\varphi_{\rho}}+2 \Rightarrow \left[\begin{array}{c}
\rho=\rho_{0}=\text { const, } \\
w=\rho_{0} e^{i \varphi}+2,|w|=\text { const } \\
\varphi=\varphi_{0}=\text { const } \\
w=\rho e^{i \rho_{0}}+2, \text { arg } w=\text { const. }
\end{array}\right.
$$

The mapping $w=3z+2$ performs the following sequence of actions:

1) triples the modulus of the complex number $z$, preserving $\arg z$
2) shifts the point obtained in step 1) by 2 units to the right along the real axis $\operatorname{Re} z$.

In Fig. 3.5, the circle $x^{2}+y^{2}=4$ and its image under the function $w$ are shown (for the point $z_{0}=\sqrt{2}+\sqrt{2} i$, the sequence of mapping is indicated by arrows

$$
\sqrt{2}+\sqrt{2} i=z_{0} \xrightarrow{w} 3z_{0}+2=3\sqrt{2}+2+3\sqrt{2} i
$$

the origins of the coordinate systems are coincident, the scales of the axes are the same):

$$
x^{2}+y^{2}=4 \stackrel{w}{\rightarrow}(x-2)^{2}+y^{2}=36 .
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-184.jpg?height=504&width=595&top_left_y=1208&top_left_x=332)

Fig. 3.5