What is the principal branch of the complex arcsine, in simple terms?

Solution 1:

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 $.