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You want to find the image of $\vert z\vert<1$ under $T(z)=\frac{iz-i}{z+1}=\frac{i(z-1)}{z+1}$. Take 3 boundary points: $1,i,-1$ and note that $T(1)=0$ $T(i)=-1$ $T(-1)=\infty$ Also observe the direction of the points, first is $1$ then $i$ and then $-1;$ and the interior is in the LHS, thus following the same direction of the points $0,-1,\infty$, ...

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Your answer is correct. Note that the inner angle of the lens at $z_1$ and $z_2$ is ${2\pi\over3}$. This angle remains after the Möbius transformation $z\mapsto w$. Only the exponent ${3\over2}$ can convert this angle to $\pi$, as required for a half plane.

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You almost have the right idea with your "diametrically opposite" approach, but not quite. Instead, we want to use the idea of symmetry with respect to a circle--see Definition 13 on page 81--to give us a third point. As mentioned after the development of $(11)$ on page 81, the symmetric point to the center of a circle is $\infty.$ Thus, by the Symmetry ...

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Moebius transformations map circles* (meaning circles or lines) to circles*. The problem tells about two circles* in the $z$-plane. Compute $f(-1)$, $f(0)$, $f(1)$, and $f(i)$ in order to obtain three points of each of the two image circles*. It turns out that $f(A)$ is one of the quadrants in the $w$-plane. In order to determine which one compute $f(2i)$ as ...

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