# Complex numbers, inverse trig and hyperbolic

Liquidxlax

## Homework Statement

edit* It says Verify the formulas in problems

arcsin(z) = -iln(iz ±sqrt(1-z^2))

arccos(z) = iln(z ±sqrt(1-z^2))

tanh-1z = (1/2)ln((1+z)/(1-z))

## The Attempt at a Solution

yeah, my prof just threw it at us, all i have is nothing... absolutely nothing. I don't know why he does this to us.

The best i can think of is

z= sin(-iln(iz ±sqrt(1-z^2)))

z = (eln(iz ±sqrt(1-z^2)) + e-ln(iz ±sqrt(1-z^2)))/2i

((iz ±sqrt(1-z^2)) + (iz ±sqrt(1-z^2)-1)/2i

but i doubt that is right

## Answers and Replies

jackmell
You have really got to get squared with the complex logarithm and stop using the ln symbol to denote it. Use ln only when referring to the real log function. Use log to refer to the complex log function.

So we know:

$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=w$$

Now, how do you solve for z in that? I'll show you some of the steps. You try to fill in the blanks if you want:

$$2iw=e^{iz}-\frac{1}{e^{iz}}$$

$$2iwe^{iz}=e^{2iz}-1$$

$$e^{iz}=\frac{2iw+\sqrt{-4w^2+4}}{2}$$

$$iz=\log(iw+\sqrt{1-w^2})$$

Go through that one, then try arccos(z) the same way.

Keep in mind that's a complex log function: $\log(z)=\ln|z|+i(\theta+2n\pi)$ and note I do not use $\pm$ for the root. In Complex Analysis, it's implied to have two values. So the arcsin function is infinitely-valued and for each one of those n-values of the log function, the square root has two different values. So if I asked what is the arcsin(z) when n=10,11, there would be four values of the arcsin function for those two values of n.

Homework Helper
Hi Liquidxlax! (have a square-root: √ and a theta: θ and try using the X2 icon just above the Reply box )
Verify the formulas in problems

arcsin(z) = -iln(iz ±sqrt(1-z^2))

eugh! :yuck: … let's keep this simple

whenever I see √(1 - z2), I immediately want to do a trig subsitution.

So forget about the LHS, and just put z = sinθ in the RHS …

what do you get? Liquidxlax
Hi Liquidxlax! (have a square-root: √ and a theta: θ and try using the X2 icon just above the Reply box )

eugh! :yuck: … let's keep this simple

whenever I see √(1 - z2), I immediately want to do a trig subsitution.

So forget about the LHS, and just put z = sinθ in the RHS …

what do you get? i don't think that will help, but i did what jack said, yet i did not get the right answer

for arcsin

(iz +- sqrt(1-z^2))^2 = Log(iz +- sqrt(1-z^2)) somehow

jackmell
i don't think that will help, but i did what jack said, yet i did not get the right answer

for arcsin

(iz +- sqrt(1-z^2))^2 = Log(iz +- sqrt(1-z^2)) somehow

$$w=\frac{e^{iz}-e^{-iz}}{2i}$$

and solving for z like I showed above, do you not get the same expressions I obtain in the above post?

Last edited:
Liquidxlax

$$w=\frac{e^{iz}-e^{-iz}}{2i}$$

and solving for z like I showed above, do you not get the same expressions I obtain in the above post?

nope... but isn't it arcsinz = w, or should i ignore the rhs and just say

arcsinz = w => z=sinw => z = (e^iw - e^-iw)/2i. I mean they're very very stict with notation and variables i can't be changing them randomly even if i do say let z = w or something, they still whine and complain.

Liquidxlax
being my typical in a rush self, i slowed down and did what you said Jackmell, and it worked just fine