exponential forms

[Hypnotoad]

I'm try to prove some trig identities and I need to know how to write the following functions in their exponential forms:

sin^{-1}z
cos^{-1}z
sinh^{-1}z
cosh^{-1}z

I know how the sine and cosine functions are expressed with exponents, but I'm not sure how that would translate to the inverses of the functions.

[Tide]

If the trig functions can be written in exponential form then the inverse functions would be expressible in logarithmic form!

E.g. Let

\sin z = \frac {e^{i z} - e^{-iz}}{2i}

then solve for e^{iz} in terms of \sin z. Once you have your expression for e^{iz} then just find the natural logarithm of both sides and you have your inverse function!

[arildno]

Just a reminder:
Since the complex logarithm is a multi-valued function (i.e, not a "usual" type of function), you must choose a branch of it to gain a "proper" inverse function.

[Hypnotoad]

If the trig functions can be written in exponential form then the inverse functions would be expressible in logarithmic form!

E.g. Let

\sin z = \frac {e^{i z} - e^{-iz}}{2i}

then solve for e^{iz} in terms of \sin z. Once you have your expression for e^{iz} then just find the natural logarithm of both sides and you have your inverse function!

Maybe I could get a little more help. If I start with
\sin z = \frac {e^{i z} - e^{-iz}}{2i}

then I take the 2i term to the left and then take the natural log of both sides I get the following:

ln2i*sinz=ln(e^{2iz}-1)-iz

ln(2i*sinz)+iz=ln(e^{2iz}-1)

-2zsinz+1=e^{2iz}

e^{iz}=(-2zsinz+1)^{\frac{1}{2}}

But now I don't see how to use that. What I am trying to do is show that \sin^{-1}z=-iln({iz+(1-z^2)^{\frac{1}{2}}})

[arildno]

Set y=sin(z).
Then:
2iye^{iz}=e^{2iz}-1
(e^{iz})^{2}-2iye^{iz}-1=0
Or:
e^{iz}=\frac{2iy\pm\sqrt{-4y^{2}+4}}{2}
Or:
z=sin^{-1}(y)=-iln(iy\pm\sqrt{1-y^{2}})

[Hypnotoad]

Set y=sin(z).
Then:
2iye^{iz}=e^{2iz}-1
(e^{iz})^{2}-2iye^{iz}-1=0
Or:
e^{iz}=\frac{2iy\pm\sqrt{-4y^{2}+4}}{2}
Or:
z=sin^{-1}(y)=-iln(iy\pm\sqrt{1-y^{2}})


That is unbelievably easy. Thanks for the help, guys.