Holomorphic functions

  • Thread starter Firepanda
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  • #1
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I'm not too sure how to show this. Perhaps if I show d(ez)/dz = ez then does this conclude that ez is holomorphic on all of C?
 
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  • #2
Dick
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Why does d/dz(e^z)=e^z show e^z is holomorphic? Or at least, why are you saying you aren't sure this shows it's holomorphic?
 
  • #3
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I suppose I should be using the cauchy riemann equations then?

I was told not to use limits here so I didn't want to use the base definition of holomorphic-ness.
 
  • #4
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Why does d/dz(e^z)=e^z show e^z is holomorphic? Or at least, why are you saying you aren't sure this shows it's holomorphic?
Because in my notes he mentioned the power series
 
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  • #5
Dick
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I suppose I should be using the cauchy riemann equations then?

I was told not to use limits here so I didn't want to use the base definition of holomorphic-ness.
Sure, use Cauchy-Riemann. That's easy enough. Then you'd want to show e^(-z) is also holomorphic. Do you know the sum of holomorphic functions is also holomorphic? If not then you could just directly show cosh(z) is holomorphic using CR.
 
  • #6
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i tried to use C-R, but I'm unable to split it up into

U(x,y) or V(x,y)

using z=x+iy into cosh(z)
 
  • #7
Dick
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i tried to use C-R, but I'm unable to split it up into

U(x,y) or V(x,y)

using z=x+iy into cosh(z)
How about e^z? Can you split that up?
 
  • #8
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How about e^z. Can you split that up?
exeiy

= ex(cosy + i.siny)

=excosy + i.exsiny

The C-R satisfy this and so it is holomorphic on all of C? so this would immediatly imply cosh(z) is entire?

And I would check if it were true for e-z as well.
 
  • #9
Dick
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exeiy

= ex(cosy + i.siny)

=excosy + i.exsiny

The C-R satisfy this and so it is holomorphic on all of C? so this would immediatly imply cosh(z) is entire?

And I would check if it were true for e-z as well.
Sure, check e^(-z) as well. It's almost the same thing. I'm a little puzzled why you can split e^(z) and e^(-z) up and not cosh(z). You already said cosh(z)=(e^z+e^(-z))/2.
 
  • #10
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Sure, check e^(-z) as well. It's almost the same thing. I'm a little puzzled why you can split e^(z) and e^(-z) up and not cosh(z). You already said cosh(z)=(e^z+e^(-z))/2.
ah ye that would make more sense if i did it directly, thanks!
 

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