Can anyone help me do the following problem. Have been trying for ages but cant prove the answer.
let z=re^ih (where h is not fixed)
show that if 0<h<(Pi/2) then mod of cosz tends to infinity as r tends to infinity.
what is the behaviour of mod cosz as r tends to infinity and h=(pi/2)
(sorry i cant find the symbols for pi!)
thanks very much
A similar question is:
(a) Assume r and theta are real.Originally Posted by edgar davids
then cos(z)=[e^(iz) + e^(-iz)]/2
Write z=r (cos(theta)+isin(theta)), then
e^(iz)=e^(r i cos(theta)) e^(- r sin(theta))
Therefore for theta in the range specified:
|e^(iz)| = |e^(- r sin(theta))|
as |e^(r i cos(theta))|=1.
|e^(-iz)| = |e^(r sin(theta))|.
|cos(z)|=|[e^(iz) + e^(-iz)]/2|
...........>= ||e^(iz)/2| - |e^(-iz)]/2||.
...........>= |e^(- r sin(theta))|/2 - |e^(r sin(theta))|/2
but as r goes to infinity the second of the terms on the right goes to infinity, while the second goes to 0.
so |cos(z)| goes to infinity.
(b) When theta=pi/2 the cos(theta) terms are all zero and the sin(theta) terms all equal 1, so:
cos(z) = [e^(- r) + e( r)]/2 , and so |cos(z)| goes to infinity as r goes to infinity.
So the real part of e^(iz) is e^(-r sin(theta)), not the cosine version. Additionally this means that |e^(iz)| = |e^(-r sin(theta))| -> 0 as r -> infinity, so it is the |e^(-iz)| term that dominates as r -> infinity. But as this term also goes to infinity, we are merely switching the dominant term and the proof goes through otherwise unaffected.