Hint: for every $a,b\in\mathbb{C}$, find those $z\in\mathbb{C}$ such that $\frac{z-a}{z-b}\in\mathbb{R}$. Then find exactly when the given fraction is positive, and when it's negative.
Edit: for any $u,w\in\mathbb{C}$ we have $\arg{\frac{u}{w}} = \arg{u}-\arg{w}$. It follows that the quotient is real iff $u,w$ have the same argument (they 'point' in the same direction) or opposite arguments. How can we apply that to the first hint?
Edit: the next part is more explicit, and I'll try to give some intuition based on your familiarity with Möbius transformation (however, if you're not done thinking about the problem, you might wish to delay reading it). As mentioned by @AndrewD.Hwang in the comments above, an analytic branch of logarithm exists in any simply connected domain not including zero. Put differently, when you think of the complex plane as the Riemann sphere (infinity as the 'north' pole), the logarithm has branch points at the poles (zero and infinity), and removing any arc connecting both poles (that arc becomes the branch cut) will yield a simply connected surface on which an analytic branch of logarithm indeed exists.
Now, $\varphi(z) = \frac{1+iz}{1-iz}$ is a Möbius transformation. This means that it's conformal on the entire Riemann sphere (if you'd like, it 'distorts' the sphere in general, but locally it behaves very similarly to shifts and rotations). Note that it maps $i\mapsto 0$, $-i\mapsto \infty$, implying that the image of any arc connecting $\pm i$ under $\varphi$ is an arc connecting $0,\infty$, and indeed $\pm i$ are the branch points of $\arctan := \log\circ\varphi$, and any arc connecting them would be a branch cut (in other words, removing such an arc would yield a domain in which an analytic branch of $\arctan$ exists).
Now, as AndrewD.Hwang also mentioned, the standard 'choice' of branch cut for logarithm is the non-positive real line. It is here that my original hint could help us, as it allows us to find the arc mapped by $\varphi$ to that cut. As I said, for any $a,b\in\mathbb{C}$ we have $\frac{z-a}{z-b}\in\mathbb{R}$ iff $(z-a),(z-b)$ point in similar or opposite directions. This happens exactly when $z$ lies on the unique line passing through $a,b$, and the quotient is negative iff $z$ lies on the segment connecting $a,b$ (because that is when the arrows from $a,b$ to $z$ point in opposite directions). More rigorously, one notes:
$$\frac{z-a}{z-b}=t \iff z-a=t(z-b) \iff (1-t)z = a - tb = (1-t)a - t(b-a)\\ \iff z = a + \frac{t}{t-1}(b-a),$$
and indeed $\frac{t}{t-1}\in(0,1)$ ($z$ lies between $a,b$) iff $t$ is non-positive.
Finally, note that
$$\varphi(z) = \frac{i(z-i)}{i(-i-z)} = -\frac{z-i}{z+i}$$
is a non-positive real exactly when $z$ lies on the imaginary axis but not between $\pm i$, i.e. on
$$\{it\mid t\in\mathbb{R}, |t|\geq 1\}.$$
(This, you see, is an arc in the Riemann sphere connecting $\pm i$, which has the 'nice' property of passing through infinity--the point on the sphere that we 'dislike' to work with. Or, if you'd like, it's the only arc connecting $\pm i$ which doesn't pass through the real line, allowing us to truly extend the familiar $\arctan$ on reals.)