Prove that $\arcsin z = \frac{\pi}{2} - \arccos z$ I have $\arccos (z) = -i\ln (z + \sqrt{z^2-1})$ and $\arcsin (z)=-i \ln(iz +\sqrt{1-z^2}).$
Now I must prove, that $\arcsin (z) = \frac{\pi}{2} - \arccos (z)$.
I get:
$$\arcsin (z)=-i\ln\left(iz+\sqrt{1-z^2}\right)=-i\ln\left(iz+\sqrt{-1(z^2-1)}\right)=-i\ln\left(iz+\sqrt{i^{2}(z^2-1)}\right)=$$
$$=-\ln\left(iz+i\sqrt{z^2-1}\right)=-i\ln\left(i(z+\sqrt{z^2-1})\right)=-i\ln -i\ln\left(z+\sqrt{z^2-1}\right)=$$
$$=-i\left(\ln|i|+i\frac{\pi}{2}\right)-i\ln\left(z+\sqrt{z^2-1}\right)=\frac{\pi}{2}+\arccos (z).$$
I don't know where I make mistake, because I have "+" not "-".
 A: Suppose the statement is true, namely,
\begin{align}
\sin^{-1}(z) = \frac{\pi}{2} - \cos^{-1}(z).
\end{align}
By taking the sine of both sides leads to
\begin{align}
\sin(\sin^{-1}(z)) &= \sin(\frac{\pi}{2} - \cos^{-1}(z)) \\
z &= \sin(\pi/2) \cos(\cos^{-1}(z)) - \cos(\pi/2) \sin(\cos^{-1}(z)) \\
z &= z.
\end{align}
This shows the identity is true.
A: lets start by defining $w to be $sin(w) = z
now, we'll use the complex identity:
sin(w) = z = (exp(iw)-exp(-iw))/2i 

define t = exp(iw) therefor, by applying this to the previous equation:
z = (t - 1/t)/2i

multiply by 2i and pass all args to the right side to get:
t^2 - (2iz) - 1 = 0

a standard quadratic equation:
t = ((2iz)+sqrt((2i*z)^2 + 4))/2
t = i*z + sqrt(1 - z^2) 

but we defined t = exp(i*w) therfor:
exp(iw) = i*z + sqrt(1 - z^2)

take the log to both sides and get that:
iw = log(i*z + sqrt(1 - z^2)) 
divide by i:
w = (1/i) * log(i*z + sqrt(1 - z^2))

but as we first defined: sin(w) = z thenw = arcsin(z)
in total:
arcsin(z) = (1/i) * log(i*z + sqrt(1 - z^2))

that proves your claim is right.
A: For $z$ real and $z<1$, draw a right triangle with hypotenuse of length $1$ and a side of length $z$.  Call the angle opposite the side of length $z$, $\alpha$ and the angle adjacent to this side, $\beta$.
Obviously $\alpha + \beta = \frac{\pi}{2}$ or $$\sin^{-1} z =\frac{\pi}{2} - \cos^{-1} z, \quad (0<z<1).$$
Since the left-hand side and the right-hand side are equal on the segment $0<z<1$, and each has an analytic continuation that is entire, they are equal for all $z$ in the complex plane.
A: My usual naive approach.
From
$\arccos (z) = -i\ln (z + \sqrt{z^2-1})
$
and
$\arcsin (z)=-i \ln(iz +\sqrt{1-z^2})
$
we get
$\begin{array}\\
\arccos (z)+\arcsin (z)
&=-i(\ln (z + \sqrt{z^2-1})+\ln(iz +\sqrt{1-z^2}))\\
&=-i(\ln ((z + \sqrt{z^2-1})(iz +\sqrt{1-z^2}))\\
&=-i(\ln ((z + i\sqrt{1-z^2})(iz +\sqrt{1-z^2}))\\
&=-i(\ln (iz^2+z\sqrt{1-z^2}(-1+1)+i(1-z^2))\\
&=-i(\ln (i))\\
&=-i(\dfrac{i\pi}{2})\\
&=\dfrac{\pi}{2}\\
\end{array}
$
Throw in a $2k\pi i$
if you want.
That will change the answer to
$\dfrac{\pi}{2}+2k\pi$.
