Inverse Hyperbolic function For real no x it is fine that 
$$\sinh^{-1}x=\ln\left(x+\sqrt{x^2+1}\right)$$
But for complex number $z$ Since there is no order on complex numbers
Is it same and why?

Do you know how $\sinh^{-1}$ is found? Let's find out:
$$y=\sinh(x)=\frac{e^x-e^{-x}}{2} ~\iff~ e^{2x}-2ye^x-1=0 ~\iff~ e^x=y\pm \sqrt{y^2+1}.$$
Now, if $x$ is real, then $e^x>0$, so $e^x$ cannot be $y-\sqrt{y^2+1}$ (as it's negative), forcing it to instead be $e^x=y+\sqrt{y^2+1}$ so that $x=\ln(y+\sqrt{y^2+1})$.
However if $x$ is not real, then both $x=\ln(y\pm\sqrt{y^2+1})$ are valid complex solutions to the equation $y=\sinh(x)$. Indeed, we haven't even addressed the fact that $e^x={\rm blah}$ has multiple complex solutions for $x$ because $e^{2\pi i}=1$.
To find a domain on which we can define an inverse, first we must pick a domain on which $\sqrt{y^2+1}$ is continuous and then restrict further to a domain on which $\ln(y+\sqrt{y^2+1})$ or $\ln(y-\sqrt{y^2+1})$ (whichever one we choose to use) is continuous for our choice of $\ln$.
