By simple, I mean something akin to $\{z: -\pi < \mathrm{Im}\, z \le \pi\}$ for the logarithm or $\{z: -\pi/2 < \mathrm{Arg}\, z \le \pi/2\}$ for the square root.
I am aware of the answer, "the range of $z \mapsto -i\,\mathrm{Log}(iz + \sqrt{1-z^2})$", where one uses the principal branch of the square root, but I have having trouble sketching that region. Clearly, it is not the $-i$ times the principal branch of the logarithm (i.e., $\{z: -\pi < \mathrm{Re}\, z \le \pi\}$), as sine is not one-to-one on that region. The trouble lies in figuring out what the principal branch of $z \mapsto iz + \sqrt{1-z^2}$ is.
Before looking at other sources, I tried determining a branch of the arcsine from the relation, $\sin(x + iy) = \sin x \cosh y + i \cos x \sinh y$. From this, I got $\{z: |\mathrm{Re} \, z| < \pi/2\} \cup \{z: \mathrm{Re}\, z=\pm \pi/2, \mathrm{Im}\,z \ge 0\}$ as a natural candidate. (I think that's a branch, isn't it? That is, I've convinced myself that sine provides a bijection from this region to the entire complex plane, but I still doubt myself a little.) But this turns out not to be the principal branch, for $\mathrm{Arcsin}(2) = \pi/2 – i\ln(2+\sqrt{3})$, with a negative imaginary part. ($\sin[\pi/2 + i\ln(2+\sqrt{3})]$ is indeed 2.) It turns out that $\mathrm{sgn \, Im\, Arcsin}\, a = -\mathrm{sgn}\,a$ when $a$ is a real number whose magnitude exceeds unity. This maintains $\mathrm{Arcsin}(-z)= -\mathrm{Arcsin}\,z$. So now I hypothesize that the principal branch is almost the one I suggested as my natural candidate above. Just replace the positive imaginary parts with their negatives when the real part is $\pi/2$. Haven't proved this, though. (Somehow, it would feel more natural to me if the signs of the real and imaginary parts matched on these boundaries.)
This comes up because, as a teacher and tutor of precalculus, I often remark in passing that arcsin(2) is undefined until you get to complex analysis (which is never for almost all of my students), where things gets really hairy. I haven't done complex analysis in years, though I was assigned to teach it to undergraduates once. Just got me wondering. I've tried very hard, but, surprisingly, I have been unable to find an answer to my question on the 'net.
Best Answer
I'm not sure that there is an answer for the principal branch; that is, I'm not sure that there is a widely accepted convention. Someone told you that $ \arcsin 2 = \pi / 2 - \mathrm i \ln ( 2 + \sqrt 3 ) $, but someone else might tell you something else. But I'll answer based on three things that you stated in your question: $ \arcsin z = - \mathrm i \ln ( \mathrm i z + \sqrt { 1 - z ^ 2 } ) $; $ z \in \operatorname { ran } ( \ln ) $ iff $ - \pi < \Im z \leq \pi $; and $ z \in \operatorname { ran } ( \surd ) $ iff $ \Re z \geq 0 $ and $ \Re z > 0 $ if $ \Im z < 0 $. Then the answer is as you were suspecting: $ z \in \operatorname { ran } ( \arcsin ) $ iff: $ - \pi / 2 \leq \Re z \leq \pi / 2 $, $ \Re z > - \pi / 2 $ if $ \Im z < 0 $, and $ \Re z < \pi / 2 $ if $ \Im z > 0 $. Here is a graph from Desmos (with $ x = \Re z $ and $ y = \Im z $):
To see that this is correct, let me show where each part comes from. First of all, if $ z = \pm 1 $, then $ \arcsin z = \pm \pi / 2 $; you know that this is true if you're working in the real numbers, but it's also what the formula gives: $ 1 - z ^ 2 = 0 $, so $ \sqrt { 1 - z ^ 2 } = 0 $, so $ \mathrm i z + \sqrt { 1 - z ^ 2 } = \pm \mathrm i $, so $ \ln ( \mathrm i z + \sqrt { 1 - z ^ 2 } ) = \pm \mathrm i \pi / 2 $, so $ - \mathrm i \ln ( \mathrm i z + \sqrt { 1 - z ^ 2 } ) = \pm \pi / 2 $. So these are the black and green dots in the picture above.
Next, if $ z $ is real with $ \lvert z \rvert > 1 $, we have $ z ^ 2 > 1 $ so that $ 1 - z ^ 2 < 0 $, so we can rewrite $ \sqrt { 1 - z ^ 2 } $ as $ \mathrm i \sqrt { z ^ 2 - 1 } $, making $ \mathrm i z + \sqrt { 1 - z ^ 2 } = \mathrm i ( z + \sqrt { 1 - z ^ 2 } ) $, a purely imaginary number. A bit of calculus or a suggestive graph will show that the values taken by $ z + \sqrt { 1 - z ^ 2 } $ when $ z > 1 $ consist of the real numbers greater than $ 1 $, so when we multiply this by $ \mathrm i $ and take the logarithm, we get $ \pi / 2 $ as the imaginary part but any positive real number as the real part. Then multiplying this logarithm by $ - \mathrm i $, we get a number whose real part is $ \pi / 2 $ and whose imaginary part is negative, so this is the purple half-line in the graph above. Similarly (but less obviously, so a graph really helps here), the values of $ z + \sqrt { 1 - z ^ 2 } $ when $ z < 1 $ consist of the real numbers strictly between $ - 1 $ and $ 0 $, so when we multiply this by $ \mathrm i $ and take the logarithm, we get $ - \pi / 2 $ as the imaginary part and any negative real number as the real part. Then multiplying this logarithm by $ - \mathrm i $, we get a number whose real part is $ - \pi / 2 $ and whose imaginary part is positive, so this is the blue half-line in the graph above.
Any other value of $ z $ gives us a value of $ \mathrm i z + \sqrt { 1 - z ^ 2 } $ with a positive real part (see How can we show that $\mathrm{Re}(z + \sqrt{1+z^2}) \ge 0$ for all complex $z$? for why), so that its logarithm has an imaginary part strictly between $ - \pi / 2 $ and $ \pi / 2 $. Then when we multiply this logarithm by $ - \mathrm i $, we get a number whose real part is strictly between $ \pi / 2 $ and $ - \pi / 2 $, so in the shaded red strip. I won't go through the work to show that all of this strip is included, but you basically think about where the boundary could be and see that it can only happen when the real part of the arcsine is approaching $ \pm \pi / 2 $.