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- Oct 23rd 2006, 11:47 PM #1BertieGuest
## Big Problem

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

- Oct 24th 2006, 10:00 AM #2

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- Oct 24th 2006, 10:00 PM #3

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A similar question is:

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.

similarly:

|e^(-iz)| = |e^(r sin(theta))|.

Now

|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.

RonL

- Oct 25th 2006, 05:57 AM #4
There is a slight problem here, but it doesn't affect the result.

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.

-Dan

- Oct 25th 2006, 09:09 AM #5

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