Discover Complete Functions for (e^z)/(z^2) with Limit 0 | Image Included

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can you elaborate on what the definition of a complete function is?
 
I'm sorry , I meant entire function
 
A function that is analytic on the entire complex plane.
 
HallsofIvy said:
A function that is analytic on the entire complex plane.
I know. So what should i do?
 
Suppose there aren't any? Is that an option? Can that be justified?
 
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jackmell said:
Suppose there aren't any? Is that an option? Can that be justified?
I don't know. I want to be sure before i write that there aren't...
 
Cosmossos said:
I don't know. I want to be sure before i write that there aren't...

What makes you think there aren't? Not just by what I said huh? That exponent in there is a non-polynomial entire function which by Picard's first theorem, grows without bounds in the complex plane. z^2 has a pole at infinity which means it becomes unbounded too. What would f have to be to tame all that down to something that is never larger than 3?

Also, I'm no expert and I'm not 100% sure about this.
 
what's the actual question as well? may help to write it explicitly

my initial thoughts agree with jackmell

f(z) must be entire, so as long as it is non-constant it must take every value in the complex plane (pretty much)

so in really from f you need a function which:
bounds the z^2 behaviour at large |z| by mult
bounds the e^z behaviour at large |z| by addition
has its behaviour at large |z| canceled by the actions of ..*z^2 ..+e^x

now one way to approach it may be to write
f(z) = u(z) + iv(z)

then maybe you can make use of the fact that if f is continuous, first derivtaives of u&v exist everywhere and satisfy Cacuhy Reimann, then f is holomorphic
 
  • #10
There's a much simpler approach. Note that if f is entire, then z^2f(z)-3+e^z is also entire. And what do you know about bounded entire functions?
 
  • #11
nice point
 
  • #12
Citan Uzuki said:
There's a much simpler approach. Note that if f is entire, then z^2f(z)-3+e^z is also entire. And what do you know about bounded entire functions?

I know it's equal to a constant (C). so f(z)=(C+3)/z^2 -(e^z)/(z^2)
then I know that for f(z) to be entire C+3=0 and then I need to find what is the value of e^z/z^2 in 0.
this is what i did in the first place...
 
  • #13
not quite, this shows that the only possible form for f(z) is
f(z) = (C+3)/z^2 -(e^z)/(z^2)

now if you choose C=-3, you get
f(z)=-(e^z)/(z^2)
which is not analytic at z=0, so this is not a good choice

are there any other choices for C that would remove this behaviour?
 
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  • #14
if so, check whether that give an entire function
if not, you've proved there is no solution
 
  • #15
lanedance said:
not quite, this shows that the only possible form for f(z) is
f(z) = (C+3)/z^2 -(e^z)/(z^2)

now if you choose C=-3, you get
f(z)=-(e^z)/(z^2)
which is not analytic at z=0, so this is not a good choice

are there any other choices for C that would remove this behaviour?

I don't see any other constant that would remove this behaviour. am i wrong?
And as I understand from this thread e^z/z^2 isn't analytic so there are no functions..
 
  • #16
i haven't work it through fully... however noting that e^{0}=1, then consider C=-2 as this gives
f(z) = (1-e^z)\frac{1}{z^2}

this is the only value that gives f a chance of reasonable limiting behaviour at z=0, though I still don't think it makes it analytic there. Expanding e^z around 0

e^z = 1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+..
1-e^z = -z-\frac{z^2}{2!}-\frac{z^3}{3!}+..
(1-e^z)\frac{1}{z^2}= -\frac{1}{z}-\frac{1}{2!}-\frac{z}{3!}-...

so the pole order is reduced, but still there - though obvious you could do this for general C to show there is no available f
 
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  • #17
Ok, thank you.
One more thing, If I have a function and its denominator zero order and the numerator zero order are the same, then we have a removable singularity?
 
  • #18
if they occur at the same z then yes, in effect it is a pole of order zero
 
  • #19
I'm talking about (1-cos(z^2))/z^4 it has 4th order zero
and (e^z-1-z)/z^2 has 2nd order zero.
then they both have a removable singularity, right?
 
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