Let $C$ be the half line $\arg(z) = \frac{\pi}{4}$. Determine the image of $C$ under the transformation. $$w = \frac{z}{z+1}$$

Determine the centre and radius of the circle from which the resulting arc is taken.

The first task is to find $z = f(w)$

$$ \begin{align} w &= \frac{z}{z-1}\\ zw+ w &= z \\ z &= \frac{w}{1-w} \\ \end{align} $$We now substitute this into the equation defining the locus of points.

$$ \begin{align} \arg(z) &= \frac{\pi}{4} \\ \arg\left(\frac{w}{1-w}\right) &= \frac{\pi}{4} \\ \arg(w) - \arg(1-w) &= \frac{\pi}{4} \\ \end{align} $$Spend a moment considering $\arg(1-w)$.

This can be rewritten as $\pi+2n\pi + arg(w-1)$. In this case, we will take $n=-1$, giving $$\arg(1-w) = -\pi + \arg(w-1)$$

Substitution of this identity gives

$$ \arg(w) - \arg(w-1) = \frac{-3\pi}{4} $$ which can be expressed as $$ \arg(w-1) - \arg(w) = \frac{3\pi}{4} $$Hence our image will be the arc of the circle, above the real axis, with endpoints $z=0$ and $z=1$, as shown in the diagram

Determining the Cartesian equation is a little involved; but using geometric means, we can arrive at an answer with some speed.

By symmetry, the centre of the circle has x-coordinate, $\frac{1}{2}$. We can also find the diameter of the circle through some applications of basic trigonometry; giving $$2R = \sqrt{\left(\frac{\frac{1}{2}}{\cos \frac{\pi}{8}}\right)^2 + \left(\frac{\frac{1}{2}}{\sin \frac{\pi}{8}}\right)^2}$$

From this, we get the centre of the circle is

$$ \left(\frac{1}{2},\frac{1}{2}\tan\frac{\pi}{8} - R\right) $$