= Solution
A <conformal map> preserves angles between intersecting regular curves, including their orientation. If $f$ is holomorphic with $f'(a)\neq0$, the image tangent vectors are $f'(a)$ times the original tangent vectors. Multiplication by this nonzero complex number gives a common positive scaling and rotation, so their oriented angle is unchanged. This proves conformality throughout the stated domain.
At a zero <derivative> this can fail: $f(z)=z^2$ at zero sends rays of arguments $0$ and $\pi/4$ to rays of arguments $0$ and $\pi/2$. The angle is doubled, so this <holomorphic function> is not conformal there.
For the disk map $C(z)=i(1+z)/(1-z)$, compute
$$
\operatorname{Im}C(z)=\frac{1-|z|^2}{|1-z|^2}>0,
\qquad C'(z)=\frac{2i}{(1-z)^2}\neq0.
$$
Its inverse is $z=(w-i)/(w+i)$; for $\operatorname{Im}w>0$, the inequality $|w-i|<|w+i|$ puts this inverse in the unit disk. Both rational maps are holomorphic on their respective domains. Thus
$$
\boxed{C:D\to\mathbb H\text{ is a conformal bijection}.}
$$
For the final region, writing $z=x+iy$ gives $|e^z|=e^x\in(0,1)$ and $\arg(e^z)=y\in(0,\pi)$. Hence its exponential image is exactly the upper half of the unit disk,
$$
V=\{w:|w|<1,\ \operatorname{Im}w>0\}.
$$
The exponential is <injective> on this strip, since two equal exponential values differ by $2\pi i$ times an integer while the imaginary parts here differ by less than $\pi$. Its <derivative> never vanishes and its inverse on $V$ is the logarithm with argument in $(0,\pi)$.
Now $L(w)=(1+w)/(1-w)$ maps $V$ conformally onto the first quadrant. Indeed
$$
\operatorname{Re}L(w)=\frac{1-|w|^2}{|1-w|^2}>0,\qquad
\operatorname{Im}L(w)=\frac{2\operatorname{Im}w}{|1-w|^2}>0,
$$
and the inverse $(s-1)/(s+1)$ has modulus below one and positive imaginary part for every first-quadrant $s$. Squaring maps that quadrant bijectively onto $\mathbb H$, with nonzero <derivative>. Therefore a required composition is
$$
\boxed{F(z)=\left(\frac{1+e^z}{1-e^z}\right)^2.}
$$
Every component is a conformal bijection between the stated domains, so the composition is one as well.
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