Calculate $\arcsin (1 + i),\> \arctan (1 - i), \>\arcsin (i),\> \arctan (2i)$.
As $arcsin (z) = -ilog [iz + (1 - z^2)^{1/2}]$, then For $ arcsin(1 + i) $ we have to $$ arcsin(1 + i) = -ilog[i (1 + i) + (1- (1 + i)^{2})^{1/2}] $$, that is,$$arcsin(1 + i) = -i log(i-1 \pm \sqrt{3})$$But I don't know how to conclude what follows. Is there another, more simplified way to do this.
we also have $ arctg (z) = \frac{1}{2} log \frac{i + z}{i-z} $
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$\begingroup$Apply $\arcsin (z) = -i \ln \left(iz + \sqrt{1 - z^2}\right)$ to
1) $\arcsin (i ) = -i \ln \left((\sqrt2 -1)e^{i2\pi k}\right) = 2\pi k -i \ln(\sqrt2-1)$
2) $\arcsin (1+i ) \\ = \arcsin (\sqrt2e^{i\frac\pi4}) = -i \ln \left(\sqrt2e^{i\frac{5\pi}4} + \sqrt[4]5e^{-\frac{i }2 \arcsin\frac2{\sqrt5} }\right) \\ = -i \ln \left(-1+\sqrt{ \frac{\sqrt5+1}2}+ i \left( 1-\sqrt{ \frac{\sqrt5-1}2}\right) \right)\\ = -i \ln \left(re^{-i \theta+i2\pi k}\right) ,\>\>\>r^2 =\sqrt5+2-2\sqrt{\sqrt5+2},\>\tan\theta =\frac{\sqrt2-\sqrt{ \sqrt5-1}}{\sqrt2-\sqrt{\sqrt5+1}}\\ =2\pi k-\arctan\frac{\sqrt2-\sqrt{ \sqrt5-1}}{\sqrt2-\sqrt{\sqrt5+1}} -\frac 12\ln \left(\sqrt5+2-2\sqrt{\sqrt5+2}\right)i $
Similarly, apply $ \arctan (z) = \frac{i }{2} \ln \frac{i + z}{i-z} $ to
1) $\arctan (2i) = -\frac\pi2 -\pi k+\frac i 2\ln3$
2) $\arctan(1-i) = -\frac\pi2 -\pi k-\frac12 \arctan2 -\frac i4\ln5$
$\endgroup$ $\begingroup$Set $w=e^{iz}$, then $z=\arcsin(1+i)$ is equivalent to the quadratic equation$$ w^2-2i(1+i)w-1=0. $$Now complete squares or apply solution formulas$$ (w+1-i)^2=1+(1-i)^2=1-2i \\~\\ e^{iz}=w=-1+i\pm\left(\sqrt{\frac{\sqrt5+1}2}-i\sqrt{\frac{\sqrt5-1}2}\right) $$
For the next one you have $$ i\tan(z)=\frac{w^2-1}{w^2+1}=i(1-i)=1+i \\~\\ 0=w^2+2+i \\~\\ e^{iz}=w=\pm\left(\sqrt{\frac{\sqrt5-2}2}-i\sqrt{\frac{\sqrt5+2}2}\right) $$etc.
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