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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?

5xum
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Milan Amrut Joshi
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2 Answers2

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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 $\ln$.

anon
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  • And one more question in case of cosh^(-1)x = \ln(x+\sqrt{x^2+1}) here why donot we take -ve sign I:e cosh^(-1)=x\pm \sqrt{x^2+1} – Milan Amrut Joshi Aug 05 '15 at 08:36
  • @Milan Your comment is a bit confused. For real $x$, the function $\sinh(x)$ is bijective so it has an inverse, $\sinh^{-1}(y)=\ln(y+\sqrt{y^2+1})$. But $\cosh$ is not one-to-one on the reals - indeed it is an even function - so it has no inverse. The two solutions for $x$ to the equation $y=\cosh x$ are $x=\ln(y\pm\sqrt{y^2-1})$. (Except of course when $y=1$, where $x=0$ is the unique solution.) – anon Aug 05 '15 at 08:49
  • I am asking why Cosh^(-1)x=\ln(x+\sqrt{x^2-1}) and why not \ln(x-\sqrt{x^2-1} – Milan Amrut Joshi Aug 05 '15 at 09:02
  • @Milan And I responded to your comment with a complete explanation of the situation. Please read it until you understand it. It's possible that you are reading some source - that you are not telling us about - which chooses $f(y)=\ln(y+\sqrt{y^2+1})$ as one of the two right inverses (meaning, it is a continuous function which satisfies $\cosh f(y)=y$), but it is not a bona fide inverse because $\cosh$ has no inverse. A similar situation happens with the square root function. The function $x^2$ (with domain $\Bbb R$) has no inverse, but $\sqrt{x}$ is a right inverse of it. – anon Aug 05 '15 at 09:17
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In the complex plane, the function $z\mapsto \sinh z$ is not bijective, and therefore has no inverse.

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