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Cauchy Riemann conditions for analyticity for all values of z.

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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??
 

Answers and Replies

  • #2
Dick
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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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  • #3
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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.
 
  • #4
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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).

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.
 
  • #5
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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.
Sorry for the mis post.
 
  • #6
jbunniii
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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)?
 
  • #7
Dick
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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.
 
  • #8
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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
 
  • #9
Dick
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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
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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...
 
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  • #11
Dick
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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
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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
Dick
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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
Dick
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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
jbunniii
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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: [itex]\cos^2(x) + \sin^2(x) = 1[/itex], [itex]\cosh^2(x) - \sinh^2(x) = 1[/itex]
 
  • #16
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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
jbunniii
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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
Dick
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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
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SOLVED!! Thanks a lot guys!! Appreciate it.
 

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