Cauchy Riemann conditions for analyticity for all values of z.

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Homework Help Overview

The discussion revolves around the Cauchy-Riemann conditions for the function sin(z) and its analyticity across the complex plane, as well as the function 1/sin(z) and its behavior at specific points.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of the Cauchy-Riemann conditions to sin(z) and question the correctness of their partial derivatives.
  • There is a discussion on the implications of the analyticity of sin(z) for the function 1/sin(z) and the conditions under which it holds.
  • Some participants express uncertainty about their calculations and seek clarification on how to approach the problem of 1/sin(z).
  • Questions arise regarding the simplification of expressions and the use of trigonometric identities in the context of the Cauchy-Riemann conditions.

Discussion Status

Participants have provided guidance on checking derivatives and suggested simplifications. There is ongoing exploration of the implications of the Cauchy-Riemann conditions for both functions, with some participants expressing doubts about their manual calculations.

Contextual Notes

Some participants note that the function 1/sin(z) may not satisfy the Cauchy-Riemann conditions at specific points where the denominator equals zero, leading to further investigation of these critical points.

timeforchg
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Homework Statement



Show that sin(z) satisfies the condition. (Stated in the title)

Homework Equations





The Attempt at a Solution



f(z) = sin (z)
= sin (x + iy)
= sin x cosh y + i cos x sinh y

thus,

u(x,y)=sin x cosh y ... v(x,y)= cos x sinh y

du/dx = cos x ... dv/dx = -sin x
du/dy = -sinh y ... dv/dy = cosh y

therefore I will compare it with the Cauchy Riemann formula.

Am I right to say it doesn't satisfies the condition??
 
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Your partial derivatives are wrong. For du/dx the cosh(y) is a constant. You should get du/dx=cos(x)cosh(y). And the derivative of sinh(y) is sinh(y) not -sinh(y).
 
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timeforchg,
Please post calculus and analysis problems in the Calculus & Beyond section, not in the Precalc section. I am moving this thread.
 
Dick said:
Your partial derivatives are wrong. For du/dx the cosh(y) is a constant. You should get du/dx=cos(x)cosh(y). And the derivative of sinh(y) is sinh(y) not -sinh(y).


thanks for notifying. I mange to solve it. and yes it satifies the condition.

What if the equation is 1/sin (z)?

I try to conjugate the equation but end up with 0.
i guess i did it wrong.

How do I start with this equation?
Thanks.
 
Mark44 said:
timeforchg,
Please post calculus and analysis problems in the Calculus & Beyond section, not in the Precalc section. I am moving this thread.

Sorry for the mis post.
 
timeforchg said:
thanks for notifying. I mange to solve it. and yes it satifies the condition.

What if the equation is 1/sin (z)?
Well, you just established that sin(z) is analytic for all z. Do you know any theorems (such as the quotient rule) that could help with 1/sin(z)?
 
timeforchg said:
thanks for notifying. I mange to solve it. and yes it satifies the condition.

What if the equation is 1/sin (z)?

I try to conjugate the equation but end up with 0.
i guess i did it wrong.

How do I start with this equation?
Thanks.

You'll have to show what you did before anyone can tell you what are doing wrong. That's the right way to start.
 
am i right to say that 1/sin (z) in (a+jb) format is

sin(x) cosh(y)/ sin^2 (x) cosh^2 (y) + cos^2 (x) sinh^2 (y) - j cos(x) sinh(y)/ sin^2 (x) cosh^2 (y) + cos^2 (x) sinh^2 (y) ??

Correct me if I'm wrong
 
timeforchg said:
am i right to say that 1/sin (z) in (a+jb) format is

sin(x) cosh(y)/ sin^2 (x) cosh^2 (y) + cos^2 (x) sinh^2 (y) - j cos(x) sinh(y)/ sin^2 (x) cosh^2 (y) + cos^2 (x) sinh^2 (y) ??

Correct me if I'm wrong

Yes, that looks right. Take a look at jbunniii's suggestion above before you wade in and start taking partial derivatives. It'll get pretty complicated. You can simplify the denominator a bit before you start, but still.
 
  • #10
Dick said:
Yes, that looks right. Take a look at jbunniii's suggestion above before you wade in and start taking partial derivatives. It'll get pretty complicated. You can simplify the denominator a bit before you start, but still.

It's nasty..

If by theorem, 1/sin(z) will satisfy Cauchy - Riemann conditions for all values of z except z = k pi + pi/2, ( k=0, +- 1, +- 2 ...), where the denominator of the function equals to zero. ??but I try it out manually, it seems it doesn't satisfy the equation. hmm...
 
Last edited:
  • #11
timeforchg said:
It's nasty..

If by theorem, 1/sin(z) will satisfy Cauchy - Riemann conditions for all values of z except z = k pi + pi/2, ( k=0, +- 1, +- 2 ...), where the denominator of the function equals to zero. ??


but I try it out manually, it seems it doesn't satisfy the equation. hmm...

Yes, it is nasty to show it directly. But your theorem tells you it should if the denominator is nonzero. If your manual work tells you not, it's wrong. Again, if you want someone to correct that you'll have to post what you've done.
 
  • #12
Dick said:
Yes, it is nasty to show it directly. But your theorem tells you it should if the denominator is nonzero. If your manual work tells you not, it's wrong. Again, if you want someone to correct that you'll have to post what you've done.


Oopss.. Ok here it goes.. Very nasty.
 

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  • #13
timeforchg said:
Oopss.. Ok here it goes.. Very nasty.

Ouch. Ok, I asked for that. Very nasty indeed. You are going to have use trig and hyperbolic trig identies to show CR holds. But we know it does. Do you really HAVE to it that way? Are you working with computer algebra system or did you do this by hand? If you doing this with software then you could at least check that if you put in x=1 and y=2 for example that CR holds. I don't think anyone will actually check a mess like that. I know I don't feel like doing it. Sorry.
 
  • #14
BTW, if you really do have to do it that way, you might think about my suggestion to try and simplify the denominator before you start differntiating using identities. You could write it as cosh(y)^2-cos(x)^2 or sinh(y)^2+sin(x)^2, I think. That might help a little, but geez.
 
  • #15
Dick said:
Ouch. Ok, I asked for that. Very nasty indeed. You are going to have use trig and hyperbolic trig identies to show CR holds. But we know it does. Do you really HAVE to it that way? Are you working with computer algebra system or did you do this by hand? If you doing this with software then you could at least check that if you put in x=1 and y=2 for example that CR holds. I don't think anyone will actually check a mess like that. I know I don't feel like doing it. Sorry.

There's some obvious low-hanging fruit in all four expressions. Each would immediately benefit from application of one of these basic identities: \cos^2(x) + \sin^2(x) = 1, \cosh^2(x) - \sinh^2(x) = 1
 
  • #16
Dick said:
Ouch. Ok, I asked for that. Very nasty indeed. You are going to have use trig and hyperbolic trig identies to show CR holds. But we know it does. Do you really HAVE to it that way? Are you working with computer algebra system or did you do this by hand? If you doing this with software then you could at least check that if you put in x=1 and y=2 for example that CR holds. I don't think anyone will actually check a mess like that. I know I don't feel like doing it. Sorry.

Hahaha.. do this by hand?? oh my god! I would suffer from brain damage. ouh well. Ok a quick question. If I need to calculate the derivative of 1/sin(z) at z=0, +- pi/2, +- pi, +-3pi/2...

I used quotient rule to get -cos (z)/ sin^2 (z)

Thus, sub the values of z into the equation. I will get

f '(0) = infinity
f '(pi/2) = 0
f '(-pi/2) = 0
f '(pi) = infinity
f '(-pi) = infinity
f '(3pi/2) = 0
f '(-3pi/2) = 0

am i approaching the right way?
 
  • #17
Also, the expressions for du/dx and dv/dy already have one term in common. So only the remaining term needs to be compared.

Similarly, dv/dx and du/dy have a term in common, although if C-R is to be satisfied, there should be a negative sign on one of them. Perhaps a differentiation error?
 
  • #18
timeforchg said:
Hahaha.. do this by hand?? oh my god! I would suffer from brain damage. ouh well. Ok a quick question. If I need to calculate the derivative of 1/sin(z) at z=0, +- pi/2, +- pi, +-3pi/2...

I used quotient rule to get -cos (z)/ sin^2 (z)

Thus, sub the values of z into the equation. I will get

f '(0) = infinity
f '(pi/2) = 0
f '(-pi/2) = 0
f '(pi) = infinity
f '(-pi) = infinity
f '(3pi/2) = 0
f '(-3pi/2) = 0

am i approaching the right way?

I'm not sure what the point of this is, but ok. Sure. I would say f'(0) does not exist instead of infinity.
 
  • #19
SOLVED! Thanks a lot guys! Appreciate it.
 

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