Complex functions

[Raees]

w = (m)*coth(z)

what is w as x → ± ∞ and |y| ≤ π/2

How would I go about doing this?

 

[pka]

w = (m)*coth(z)
what is w as x → ± ∞ and |y| ≤ π/2
How would I go about doing this?

Can you tell us what \(\displaystyle (m)*\) is meant to be?
Is \(\displaystyle \coth(z)\) the complex hyperbolic cotangent function?
If so, is the imaginary part restricted to \(\displaystyle |y|\le \frac{\pi}{2}\) while \(\displaystyle x\to\pm\infty\)?

 

[Dr.Peterson]

What is coth(x+iy)?

 

[Raees]

Can you tell us what \(\displaystyle (m)*\) is meant to be?
Is \(\displaystyle \coth(z)\) the complex hyperbolic cotangent function?
If so, is the imaginary part restricted to \(\displaystyle |y|\le \frac{\pi}{2}\) while \(\displaystyle x\to\pm\infty\)?

m is just a real number constant.
Yes, it is the complex hyperbolic function.
Yes, that’s right about the restrictions.

Thanks

 

[Raees]

What is coth(x+iy)?

coth(x+iy) = (e^(2x)*e^(2iy) + 1)/(e^(2x)*e^(2iy) - 1)

Where would I go from here? I can do the limit as x→±∞. but y has a range of values, |y| ≤ π/2, so how would I approach the y value?

Thanks.

 

[HallsofIvy]

I would be inclined to divide both numerator and denominator by e^(2x) to get (e^(2iy)+ e^(-2x))/(e^(2iy)- e^(-2x)). As x goes to infinity (your problem says x→±∞ but in the first quadrant x can only go to +infinity) e^(-2x) goes to 0.