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Background: This is Brown, Churchill, Complex Variables ch 4,sec42, prob 9

My attempt: $$\int_{-2}^{2}(\sqrt{4-y^2}-iy)\overset{\frac{dz}{dy}}{\overbrace{\left (\frac{-y}{\sqrt{4-y^2}}+i\right )}}dy$$ $$\int_{-2}^{2}i\left (\frac{y^2}{\sqrt{4-y^2}}+\sqrt{4-y^2} \right ) dy $$ $$\int_{-2}^{2}i\left (\frac{4}{\sqrt{4-y^2}}\right )dy\tag{1}$$ $$\int_{-2}^{2}\left (\frac{-4}{\sqrt{y^2-4}}\right )dy$$ $$-4\cosh^{-1}(y/2)|_{-2}^2$$ Now aside from arccos having a branch cut along the real axis $(-\infty,1)$ making it undefined, what did I do wrong if anything. (Noting at line 1 integrating as $\left[ 4i\sin^{-1}(y/2)\right ]_{-2}^{2}$ gives $4\pi i$. I've tried using the definition to see if the answers are interchangeable along some branch cut.
$$-\cosh^{-1}(z)=-\ln(z+(z^2-1)^{1/2})$$ $$=-\ln(z+|z^2-1|^{1/2}e^{i/2\arg(z^2-1)})=-\ln(z+|z^2-1|^{1/2}e^{i/2[\arg(1-z^2)+\pi]})$$ $$=-\ln(z+i|1-z^2|^{1/2}e^{i/2\arg(1-z^2)})=-\ln(z+i(1-z^2)^{1/2})$$

Which is starting to look like $\sin^{-1}(z)=-i\ln[iz+(1-z^2)^{1/2}]$ I may be one reciprocal and power of i away if someone can help me.

On a related note, how could I convert an answer when integrating $$\frac{d}{dz}\cos^{-1}(z)=\frac{-1}{(1-z^2)^{1/2}}$$ $$\frac{d}{dz}\sin^{-1}(z)=\frac{1}{(1-z^2)^{1/2}}$$

What shows $-1\cdot\cos^{-1}(z)=i\ln[z+i(1-z^2)^{1/2}]=\sin^{-1}=-i\ln[iz+(1-z^2)^{1/2}]$? I'm interested in a link or calculation.

rubik
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user5389726598465
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    Notice that $\int_{-2} ^2 \Bbb i \frac 4 {\sqrt {4-y^2}} \ \Bbb dy$ can be computed directly as $4 \Bbb i \arcsin \frac y 2 \Big| _{-2} ^2 = 4 \pi \Bbb i$, without the need to play with $\Bbb i$ and hyperbolic functions. – Alex M. Aug 02 '16 at 06:44
  • So, C is the circle with centre O and radius 2 ? – Jean Marie Aug 02 '16 at 06:46
  • @JeanMarie: No, just the right half-circle. – Alex M. Aug 02 '16 at 06:47
  • yes, evaluating at (1) with arcsin is much easier, but I was wondering why the way I tried may not work or be right, so that I don't guess when solving. There's a reason for everything. – user5389726598465 Aug 02 '16 at 06:48
  • and the question specifically states you can't use $2e^{i\theta}$ to travel the circle around the right half (see wording). – user5389726598465 Aug 02 '16 at 06:49
  • A simple reason for this pitfall is that change of variable $y \rightarrow iy$ is not allowed in (Riemann) integrals. – Jean Marie Aug 02 '16 at 06:50
  • @JeanMarie: There is no such change of variable there, take a look again. – Alex M. Aug 02 '16 at 06:51
  • Passing from $\int_{-2}^{2}i\left (\frac{4}{\sqrt{4-y^2}}\right )dy\tag{1}$ to the next line is equivalent to the unallowed change of variable $y \rightarrow iy$. – Jean Marie Aug 02 '16 at 06:55
  • @JeanMarie: No, passing from one line to the other is just taking a $-1$ out of the square root in the denominator. The change of variable that you speak about would have transformed that denominator into $\sqrt {4 + y^2}$ (notice the "plus" sign). – Alex M. Aug 02 '16 at 06:57
  • Indeed ! You are completely right, I did a mistake. But now I see another reason. In the integral $\int_{-2}^{2}\left (\frac{-4}{\sqrt{y^2-4}}\right )dy$, the function which is under the integral sign (the "integrand") is not defined between the bounds [-2,2], thus the following line cannot be written... – Jean Marie Aug 02 '16 at 08:08
  • I think in complex analysis that's allowed, otherwise i wouldn't be allowed when integrating along the y-axis either. – user5389726598465 Aug 02 '16 at 09:14
  • @user135711: You suspect correctly that it's a matter of branch cuts. The usual choice is along $(-\infty,1]$, which makes $\cosh ^{-1} (-1)$ problematic. You can find all the needed information on Wikipedia, on MathWorld and on the NIST site. – Alex M. Aug 02 '16 at 16:35

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