The question asks to
1) Find a Möbius Transformation $T(z)$ such that the region bounded by the arcs $oac$ and $obc$ is transformed to a sector (of form {z : $\theta_1$< arg(z) < $\theta_2$ }). With origin being mapped to the origin, the arc $oac$ being mapped to the positive $\Bbb R$ axis.
2) Describe the image of region $G$ under $T(z)$ found in question 1.
Feel free to use the fact that $b$ = $\omega$ = $e^{i\pi/3}$
1) So, I have found a transformation $T(z)=\frac {z(-\frac 12 - \frac {i\sqrt3}{2})} {z-\frac 32-\frac {i\sqrt3}{2}}$ by sending ($a$ to 1) ($0$ to 0) and ($c$ to $\infty$).
Using this, the region is Q1 is mapped to the sector
$[w:0<arg(w)<\frac{2\pi}{3}]$
2) The region enclosed by disc $C_1$ is $T(C_1)$ which I found by replacing the $z$ with what $z$ is in terms of $w$
If $z=\frac {z(-\frac 12 - \frac {i\sqrt3}{2})} {z-\frac 32-\frac {i\sqrt3}{2}}$ then $w=\frac {w(\frac 32 + \frac {i\sqrt3}{2})} {w+\frac 12+\frac {i\sqrt3}{2}}$
So, $T(C_1)$ becomes the upper half plane after some algebra. If $w=u+iv$ then $T(C_1)=[w: Im(w)\ge 0]$
Region enclose by $C_3$ becomes $T(C_3)$ which is a disc with center $(\frac 14,\frac{\sqrt3}{4})$ and radius $\frac {\sqrt3}{2}$
$C_2$ gets mapped to an inequality but we dont need to find all of it because all we already have the intersection of $C_1$ and $C_2$ which is the region $Oacb$ and we know that that becomes the sector mentioned in Q1
All we need to do now to find $T(G)$ is to subtract the region that the sector in Q1 gets mapped to and the sector $Obd$ from $T(C_1)$
We get $[w:Im(w)\ge 0] \setminus [[w:0<arg(w)<\frac{2\pi}{3}]\cup[w=u+iv: |w-\frac 14-\frac{\sqrt3}{4}|\le\frac34, s.t Re(w)=-\frac12]] $
Which looks like the sector in the image that I've attached
The shaded region is $T(G)$