[College Maths] - Roots of a complex hyperbolic function

[Artic_counter]

Hi guys, it's me again. I'm doing some drill right now and I've encountered a problem which is giving me some difficulties:

Find all complex roots for : sinh(z) = i

If you could show me how to get them it would be much appreciated!

[PaperclipGames]

Simply take the inverse hyperbolic sine from both sides. Since sinh-1(i) = i*pi/2 we have that z = i*pi/2. Add a factor 2i*n*pi (n a whole number) for all solutions, since sinh is periodic with period 2i*pi.

[Artic_counter]

Thanks!

But how do you determine that sinh-1(i) = i*pi/2 ?
What if it was sinh-1(1)?

[smartdude1212]

Without getting into details about prinicpal values, it's worth knowing that:

arcsinh(z) = iarcsin(-iz)

[Zapmeister]

just expand out the sinh and solve as a quadratic for e^z

(e^z - e^-z)/2 = i
=> e^2z - 2ie^z - 1 = 0
=> (e^z - i)(e^z - i) = 0
=> e^z = i
=> z = i*pi/2 + 2ni*pi

[Arkuski]

Originally Posted by Zapmeister

just expand out the sinh and solve as a quadratic for e^z

(e^z - e^-z)/2 = i
=> e^2z - 2ie^z - 1 = 0
=> (e^z - i)(e^z - i) = 0
=> e^z = i
=> z = i*pi/2 + 2ni*pi

gg Zap I did it your way

[___________]

Expand the sinh(), and then divide by i. The LHS should remind you of the formula for the imaginary part of a complex number.