Inverse hyperbolic sine of a complex variable

Question

show that :

$$sinh^{-1}(z) =ln(z+\sqrt{1+z^2})$$

my try :

I just computed $sinh(sinh^{-1}(z))$

$sinh(sinh^{-1}(z)) = \frac{e^{ln(z+\sqrt{1+z^2})}-e^{-ln(z+\sqrt{1+z^2})}}{2} = \frac{z+\sqrt{1+z^2}-\frac{1}{z+\sqrt{1+z^2}}}{2} = \frac{2z^2+2z\sqrt{1+z^2}}{2(z+\sqrt{1+z^2})} = z$

but I feel like this is not sufficient to conclude that $$sinh^{-1}(z) =ln(z+\sqrt{1+z^2})$$

I'm actually stuck on this problem. help me solve this problem please.

Answer 1

Hint:

from $$y=\sinh(z)=\frac{e^z-e^{-z}}{2}=\frac{e^{2z}-1}{2e^z}$$ we have: $$ e^{2z}-2ye^z+1=0 $$ with $e^z=x$ solve $x^2-2yx+1=0$, paying attention to the acceptability of the solutions.