Conformal mapping from a wedge to the upper half complex plane

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I found (L. J. Laslett, "On Intensity Limitations Imposed by Transverse Space-Charge Effects in Circular Particle Accelerators", Proceedings of The 1963 Summer Study on Storage Rings, Accelerators and Experimentation at Super-High Energies, Lawrence Radiation Laboratory, 1963 pp. 324-367) a conformal mapping formula which is supposed to map a wedge (coordinate $z=-X$) with an angle of $2\alpha$ at the sharp corner into the upper half plane. The sharp corner of the wedge is mapped to $z=0$.

$$z' =ic'\left(\frac{z}{X} +1\right)^{\frac{\pi}{2\alpha}}$$

Does this conformal transformation indeed map a wedge on the upper half plane?
If I parametrize the points on the upper flank of the wedge by (I represent here the complex numbers as vector $ z=x+iy = (x,y)^T$):

$$z = \left(\begin{array}{c} -X \\ 0\end{array}\right) + t\left(\begin{array}{c} 1 \\ \tan(\alpha)\end{array}\right)$$

with $t=[0, \infty]$ I have difficulties to see that $z'$ ends up to be a real number -- i.e. the lower limit of the upper plane.

Moreover I wonder of the role of the parameter $c'$. Is it free or does it depend on $X$ or $\alpha$?

Thank you for any help.

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"Factoring" complex mappings into mappings whose geometric effect is easy to see often clarifies.

Here, if I understand the notation, $z$ lies in a wedge $D$ with vertex $-X$ and interior angle $2\alpha$, but with arbitrary orientation. Then:

The mapping $z \mapsto z/X$ scales $D$ to a wedge with vertex $-1$ and interior angle $2\alpha$.

The mapping $z \mapsto w := z/X + 1$ maps $D$ to a wedge $D'$ with vertex $0$ and interior angle $2\alpha$.

The mapping $w \mapsto w^{\pi/(2\alpha)}$ (for some fixed branch of the complex exponential) sends the wedge $D'$ to some half-plane $H'$ through $0$, i.e., to a wedge with interior angle $\pi$. The important geometric features are, this mapping is conformal and invertible in the open wedge, and angles at $0$ are multiplied by $\pi/(2\alpha)$.

Consequently, the composite mapping $z \mapsto (z/X + 1)^{\pi/(2\alpha)}$ maps $D$ to the half-plane $H'$, conformally and invertibly on the interior of $D$.

Context now suggests the constant $c'$ is chosen so that multiplying by $ic'$ rotates the half-plane $H'$ to the upper half-plane $H$. Presumably, calling the constant $ic'$ rather than $c'$ simplifies matters elsewhere in the paper, or is standard in the literature.