# Exponential forms

1. Oct 19, 2004

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.

2. Oct 19, 2004

### 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!

3. Oct 19, 2004

### 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.

4. Oct 19, 2004

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 $$\sin^{-1}z=-iln({iz+(1-z^2)^{\frac{1}{2}}})$$

5. Oct 19, 2004

### 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:
$$z=sin^{-1}(y)=-iln(iy\pm\sqrt{1-y^{2}})$$

6. Oct 19, 2004