TEAM PROJECT. We map $0 \mapsto r_{0}, 2 c \mapsto-r_{0}$, obtaining from (2) with $b=z_{0}$ the conditions
$$r_{0}=-\frac{-z_{0}}{-1}=z_{0}, \quad-r_{0}=\frac{2 c-z_{0}}{2 z_{0} c-1}=\frac{2 c-r_{0}}{2 r_{0} c-1},$$
hence
$$r_{0}=\frac{1}{2 c}\left(1-\sqrt{1-4 c^{2}}\right)$$
$r_{0}$ is real for positive $c \leqq \frac{1}{2}$. Note that with increasing $c$ the image (an annulus) becomes slimmer and slimmer.

$$\Phi=100(1-(1 / \pi) \operatorname{Arg}(z-1)), F(z)=100(1+(i / \pi) \operatorname{Ln}(z-1))$$
$\pm i$ are fixed points, and straight lines are mapped onto circles (or straight lines). From this the assertion follows. (It also follows by setting $x=0$ and calculating $|w| .$ )
The function $z=Z^{2}$ maps the first quarter of $|Z| \leqq 1$ onto the upper half of the unit disk $|z| \cong 1$, the segments $0 \leqq X \leqq 1$ and $0 \leqq Y \leqq 1$ being mapped into the $x$-axis, where the potential is zero (Fig. 372a). From this the result follows.

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