Conformally map Swiss cross domain to a half-plane?

Question

Let $\Omega$ be the complex plane with the real interval $[-1,1]$ removed and with the purely imaginary interval $[-i,i]$ removed. (This domain has appeared at least once before: Conformal Map Question .)

I am comfortable with the following tool-kit: $\log$ and $\exp$; Linear fractional transformation a.k.a Möbius map; $z \mapsto z^r$ with $r>0$ and rational; the map $z \mapsto z + \frac{1}{z}$; the map $z \mapsto \sin z.$ With these tools I can solve most simple conformal mapping problems.

I have come up with eight steps which when composed, map $\Omega$ onto a half-plane; see below.

My question is, can I stream-line my solution to perhaps use fewer primitive steps? I'd like to know if my tool-kit is missing a useful tool.

Or if my toolkit is good enough, can I avoid using $z \mapsto z + \frac{1}{z}$ and the square root each one repeatedly in my eight steps? Maybe the order in which I apply my primitive maps is not the best. (Also if you spot an error in my solution, please point it out.)

0. The starting point is the unit "plus sign" or "Swiss cross".

1. By applying $z \mapsto \frac{az+b}{cz+d}=\frac{i-z}{z+i},$ the cross becomes an infinite fork; $[-1,1]$ becomes the right unit half-circumference, while $[-i,i]$ becomes the positive real axis.

2. We map the fork to a frog face via a square root, choosing the branch $\{0<\theta<2\pi\}\mapsto\{0<\frac{\theta}{2}<\pi\}.$

3. Now apply $z \mapsto z + \frac{1}{z}$; the boundary is now the real axis, excepting the interval $(-\sqrt 2,\sqrt 2)$.

4. Next we stretch and shift with $z \mapsto \frac{z-\sqrt2}{2 \sqrt 2}$; the boundary is now the real axis, excepting the interval $(-1,0)$.

5. Apply the same branch of square root used already in step 2; now we have the upper half-plane with the ray $[i,\infty)=\{x=0,y \geq 1\}$ removed.

6. Now use the linear fractional map $z \mapsto \frac{z-i}{z+i};$ this gives us the unit disc with $[0,1)$ removed.

7. Apply the same branch of square root already used twice; now we have the upper half-unit disc.

8. Last of all apply $z \mapsto z + \frac{1}{z}$; we obtain the lower half-plane $-\mathbb{H}.$