Mathematics 79 Online
OpenStudy (anonymous):

Let "a" be a fixed real number greater than 1. If \[cosh ^{−1}(x+iy)+cosh^{−1}(x−iy)=cosh^{−1}a\] , then prove that (x,y) lies on an ellipse. Let "a" be a fixed real number greater than 1. If \[cosh ^{−1}(x+iy)+cosh^{−1}(x−iy)=cosh^{−1}a\] , then prove that (x,y) lies on an ellipse. @Mathematics

OpenStudy (anonymous):

OpenStudy (jamesj):

What is arccosh as a log function? Write it out explicitly.

OpenStudy (jamesj):

I.e., if b = cosh(a) = 1/2 (e^a + e^-a), then a = what as a function of b?

OpenStudy (jamesj):

Once you write out cosh^-1 = arccosh explicitly as the ln(something), this will become clearer.

OpenStudy (anonymous):

a=arccosh(b)

OpenStudy (jamesj):

Yes, but turn that into a log function. I.e., if b = cosh(a) = 1/2 (e^a + e^-a) then (e^a)^2 - 2b e^a + 1 = 0 Hence e^a = b +- sqrt(b^2 - 1); now it has to be the positive root i.e., a = ln (b + sqrt(b^2 -1) ) Hence arccosh(x + iy) = ln ( ..... ) arccosh(x - iy) = ln( .... )

OpenStudy (anonymous):

Actually I am new to this thing, and I am not still aware of this relation with log, which you are talking of. Can you once mention it?

OpenStudy (jamesj):

I just derived the expression immediately above. If b = cosh(a), then a = ln(b + sqrt(b^2 - 1))

OpenStudy (anonymous):

Ok it will take a little time for me to get that thing settled in my brain. May I get you back on this post, once I need somthing else?

OpenStudy (jamesj):

Sure. I may or may not be around.

OpenStudy (anonymous):

OK, I will keep posting here.