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 isz= sin(-iln(iz ±sqrt(1-z^2)))

z = (e^{ln(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

 

[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:

[tex]\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=w[/tex]

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:

[tex]2iw=e^{iz}-\frac{1}{e^{iz}}[/tex]

[tex]2iwe^{iz}=e^{2iz}-1[/tex]

[tex]e^{iz}=\frac{2iw+\sqrt{-4w^2+4}}{2}[/tex]

[tex]iz=\log(iw+\sqrt{1-w^2})[/tex]

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

Keep in mind that's a complex log function: [itex]\log(z)=\ln|z|+i(\theta+2n\pi)[/itex] and note I do not use [itex]\pm[/itex] 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.

 

[tiny-tim]

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! … 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! … 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 answerfor 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 answerfor arcsin

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

Tell you what, start with:

[tex]w=\frac{e^{iz}-e^{-iz}}{2i}[/tex]

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

 

[Liquidxlax]

Tell you what, start with:

[tex]w=\frac{e^{iz}-e^{-iz}}{2i}[/tex]

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