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

[Cosmossos]

Hello
I need to find all the complete function that satisfy:
http://img153.imageshack.us/img153/7717/unledkmb.jpg

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I got stuck in finding the limit in 0 of the expression (e^z)/(z^2)

Please help me

 

[lanedance]

can you elaborate on what the definition of a complete function is?

 

[Cosmossos]

I'm sorry , I meant entire function

 

[HallsofIvy]

A function that is analytic on the entire complex plane.

 

[Cosmossos]

A function that is analytic on the entire complex plane.

I know. So what should i do?

 

[jackmell]

Suppose there aren't any? Is that an option? Can that be justified?

 

[Cosmossos]

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...

 

[jackmell]

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.

 

[lanedance]

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

 

[Citan Uzuki]

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?

 

[lanedance]

nice point

 

[Cosmossos]

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...

 

[lanedance]

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?

 

[lanedance]

if so, check whether that give an entire function
if not, you've proved there is no solution

 

[Cosmossos]

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..

 

[lanedance]

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

 

[Cosmossos]

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?

 

[lanedance]

if they occur at the same z then yes, in effect it is a pole of order zero

 

[Cosmossos]

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?