# Big Problem

• Oct 23rd 2006, 10:47 PM
Bertie
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, 09:00 AM
CaptainBlack
Quote:

Originally Posted by Bertie
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

What you need to know is that for a general complex number:

cos(z) = [e(iz)+e(-iz)]/2.

Write out your z in Cartesian form (like a+ib) and substitute it in this

RonL
• Oct 24th 2006, 09:00 PM
CaptainBlack
A similar question is:

Quote:

Originally Posted by edgar davids
sorry to keep hassling you all the time but youre the one who seems to be the most helpful and the mostable to answer all my problems!

can you give me a hand with part a and b of this question please.

Edgar

Let z = re^itheta, where theta is fixed.

(a) Show that if 0 < theta< pi/2, then | cos z| tends to infinity as r tends to infinity.
(b) Describe the behaviour of | cos z| as r tends to infinity when theta = pi/2

(a) Assume r and theta are real.

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, 04:57 AM
topsquark
Quote:

Originally Posted by CaptainBlack
A similar question is:

(a) Assume r and theta are real.

then cos(z)=[e^(iz) + e^(-iz)]/2

Write z=r (cos(theta)+isin(theta)), then

e^(iz)=e^(r cos(theta)) e^(r i sin(theta))

Therefore for theta in the range specified:

|e^(iz)| = |e^(r cos(theta))|

There is a slight problem here, but it doesn't affect the result.

$e^{iz} = e^{ir \, cos \theta} e^{i^2r \, sin \theta} = e^{-r sin \theta}e^{ir \, cos \theta}$

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, 08:09 AM
CaptainBlack
Quote:

Originally Posted by topsquark
There is a slight problem here, but it doesn't affect the result.

$e^{iz} = e^{ir \, cos \theta} e^{i^2r \, sin \theta} = e^{-r sin \theta}e^{ir \, cos \theta}$

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

Opps

But it does change the second part, which I have adjusted accordingly.
The attachment shows the result of a numerical experiment to check the
result.

RonL