I am working on the following problem
Let $C_0$ be the circle. that passes through the points 1 and -1 and has center $c_0=ia$.
(a) Find the equation of the circle $C_0$
(b) Show that the image of the circle $C_0$ under $w=\frac{z-1}{z+1}$ is a line $L_0$ that passes through the origin.
(c) Show that the line $L_0$ is inclined at the angle $\alpha_0=\frac{\pi}{2}-Arctan(a)$
I found the equations for $C_0$ to be $|z-ia|=1+a^2$ But I am stuck on (b), since $C_0$ can not contain the origin by definition then it must map the circle $C_0$ to another circle and this circle contains the origin since $w(1)=0$. But $w(-1)=\infty$ so the image should be a vertical line, but the angle can't be constant due to (c). I'm not sure where I messed up in my logic.
You did part (b) correctly (assuming you're including lines as circles). Your statement that "the line should be vertical" is wrong, though. Just because a line goes through the origin and complex infinity doesn't mean the line is vertical! Why would it? This should be counter to an intuition of symmetry - complex infinity isn't "out there" in just one particular direction from the origin, it's "out there" in all directions simultaneously. Think about it on the Riemann sphere with north and south poles.
For (c), since Mobius transformations preserve angles, it suffices to find the angle between $C_0$ and the real axis (which is preserved by the Cayley transformation) at $z=1$.