What is the singularity of e^(-1/z^2)?

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In summary, the conversation discusses the concept of a singularity, specifically in the context of a function not being analytic at a certain point. The function in question, e^-1/z^2, has an essential singularity at z=0, as this is where the denominator goes to 0. However, the book answer given is B, which may be a typo as the conversation agrees that there is indeed a singularity at z=0. Further discussion explores the properties of this singularity and concludes that it is a non-isolated singularity, meaning it is the only singularity in the function. The correct answer should be C.
  • #1
jaus tail
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Homework Statement


upload_2018-2-2_11-1-13.png


Homework Equations


Singularity is where function is not analytic means we can't differentiate denominator. It tends to 0.

The Attempt at a Solution


expanding e^-1/z^2 = 1 - 1/z^2 + 1/z^4 - ...
This has essential singularity at z = 0 as that's where the denominator goes to 0.
but book answer is B. How?
 

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  • #2
jaus tail said:

Homework Statement


View attachment 219527

Homework Equations


Singularity is where function is not analytic means we can't differentiate denominator. It tends to 0.

The Attempt at a Solution


expanding e^-1/z^2 = 1 - 1/z^2 + 1/z^4 - ...
This has essential singularity at z = 0 as that's where the denominator goes to 0.
but book answer is B. How?

I think that is a typo. You are correct.
 
  • #3
jaus tail said:

Homework Statement


View attachment 219527

Homework Equations


Singularity is where function is not analytic means we can't differentiate denominator. It tends to 0.

The Attempt at a Solution


expanding e^-1/z^2 = 1 - 1/z^2 + 1/z^4 - ...
This has essential singularity at z = 0 as that's where the denominator goes to 0.
but book answer is B. How?
What happens at ##z=0##? And why shouldn't we be able to differentiate it there as often as we like? The graph is pretty flat in a neighborhood of the origin.
 
  • #4
fresh_42 said:
What happens at ##z=0##? And why shouldn't we be able to differentiate it there as often as we like? The graph is pretty flat in a neighborhood of the origin.

And what happens if ##z=it## and you let ##t## go to zero? The language of the problem seems to imply we should think of it as a complex function.
 
  • #5
Dick said:
And what happens if ##z=it## and you let ##t## go to zero? The language of the problem seems to imply we should think of it as a complex function.
Yes, but it hasn't been mentioned. A good example why section ##2## in the template is as important as section ##3##. A complex variable is a hidden assumption.
 
  • #6
I guess there is no singularity. For singularity e^(-1/z^2) has to go to infinite. That happens when
upload_2018-2-2_19-14-11.png

So there is no singularity. Book answer is right.
 

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  • #7
jaus tail said:
I guess there is no singularity. For singularity e^(-1/z^2) has to go to infinite. That happens when
View attachment 219559
So there is no singularity. Book answer is right.

Wrong, a denominator doesn't have to be zero for a quantity to approach infinity. If ##f(z)=e^\frac{-1}{z^2}## what is ##f(i/10)##?
 
  • #8
jaus tail said:
I guess there is no singularity. For singularity e^(-1/z^2) has to go to infinite. That happens when
So there is no singularity. Book answer is right.
Is it a real or a complex function? If complex, what about @Dick's suggestion ##z=it## with ##t \in \mathbb{R}\, , \,t \to 0\,?##
 
  • #9
Dick said:
Wrong, a denominator doesn't have to be zero for a quantity to approach infinity. If ##f(z)=e^\frac{-1}{z^2}## what is ##f(i/10)##?
That's e^(100)
Yeah you're right.
So how to solve this question?
 
  • #10
jaus tail said:
That's e^(100)
Yeah you're right.
Can somebody please tell me what the right answer is?

Didn't you already say that it has an essential singularity? I can't find any reason to disagree.
 
  • #11
But book answer is B.
upload_2018-2-2_11-1-13-png.png
 

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  • #12
jaus tail said:
But book answer is B.
Again, is it ##f\, : \,\mathbb{R} \longrightarrow \mathbb{R}## or is it ##f\, : \,\mathbb{C} \longrightarrow \mathbb{C}## with ##f(z)=\exp (-\frac{1}{z^2})\,?##
 
  • #13
Also when z = 0, then e^(-1/0) = e^(-inf) = 1/[e^inf] = 1/inf = 0. This should be infinite if this is a singularity, right?
 
  • #14
fresh_42 said:
Again, is it ##f\, : \,\mathbb{R} \longrightarrow \mathbb{R}## or is it ##f\, : \,\mathbb{C} \longrightarrow \mathbb{C}## with ##f(z)=\exp (-\frac{1}{z^2})\,?##
I'm sorry I didn't understand this.
 
  • #15
jaus tail said:
Also when z = 0, then e^(-1/0) = e^(-inf) = 1/[e^inf] = 1/inf = 0. This should be infinite if this is a singularity, right?
There is a fundamental difference whether you approach the point ##z=0## along the real number line or the imaginary number line!
 
  • #16
jaus tail said:
Also when z = 0, then e^(-1/0) = e^(-inf) = 1/[e^inf] = 1/inf = 0. This should be infinite if this is a singularity, right?

No. "inf" is not a number. The function isn't ##0## at ##z=0##, it's undefined. There's a big difference.
 
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  • #17
This is getting more and more complicated.
From what I know,
singularity means function fails to be analytic at the point.
Analytic has 2 conditions:
Ux = Vy
and
Vx = -Uy
Ux is derivative w.r.t x
Uv is derivate w.r.t y
Z = U + iV
Short cut to find out analytic is see if function is going to infinite. If it goes infinite it means function is not analytic at that point.
 
  • #18
jaus tail said:
This is getting more and more complicated.
From what I know,
singularity means function fails to be analytic at the point.
Analytic has 2 conditions:
Ux = Vy
and
Vx = -Uy
Ux is derivative w.r.t x
Uv is derivate w.r.t y
Z = U + iV
Short cut to find out analytic is see if function is going to infinite. If it goes infinite it means function is not analytic at that point.
See post #4.
 
  • #19
z = it as t tend to 0
Means e^(+1/t^2) at t tend to 0
means e(1/0) means e^(inf) means inf.
So there is a singularity at t or z = 0
But book answer is B
So can i assume that book answer is wrong?
 
  • #20
jaus tail said:
z = it as t tend to 0
Means e^(+1/t^2) at t tend to 0
means e(1/0) means e^(inf) means inf.
So there is a singularity at t or z = 0
But book answer is B
So can i assume that book answer is wrong?

Don't say things like e^(1/0) and e^(inf). That's sloppy. And it's not even true. You know the limit is zero if ##z=t## where ##t## is real. The limit is UNDEFINED. But yes, the book is wrong.
 
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  • #21
Should answer be C or D?
I think it's C. Non-isolated singularity means like there are many other singularities neighboring 0 but that's not the case here. At z = 1, the f(z) becomes 0.368.
 
  • #22
jaus tail said:
Should answer be C or D?
I think it's C. Non-isolated singularity means like there are many other singularities neighboring 0 but that's not the case here. At z = 1, the f(z) becomes 0.368.

Yes, the function is well behaved except at the single point ##z=0##. It's C.
 
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  • #23
Wow thanks. I hope this question comes in exam now. Direct 10 marks, ten such questions. :D
 
  • #24
jaus tail said:

Homework Statement


View attachment 219527

Homework Equations


Singularity is where function is not analytic means we can't differentiate denominator. It tends to 0.

The Attempt at a Solution


expanding e^-1/z^2 = 1 - 1/z^2 + 1/z^4 - ...
This has essential singularity at z = 0 as that's where the denominator goes to 0.
but book answer is B. How?
e^(-1/z^2)=0 has only zero of order 2 at z=0,and this function has no pole at z=0 that means no singularity and this further implies no isolated essential singularity
 
  • #25
jaus tail said:

Homework Statement


View attachment 219527

Homework Equations


Singularity is where function is not analytic means we can't differentiate denominator. It tends to 0.

The Attempt at a Solution


expanding e^-1/z^2 = 1 - 1/z^2 + 1/z^4 - ...
This has essential singularity at z = 0 as that's where the denominator goes to 0.
but book answer is B. How?
e^(-1/z^2)=0 has only zero of order 2 at z=0,and this function has no pole at z=0 that means no singularity and this further implies no isolated essential singularity
jaus tail said:

Homework Statement


View attachment 219527

Homework Equations


Singularity is where function is not analytic means we can't differentiate denominator. It tends to 0.

The Attempt at a Solution


expanding e^-1/z^2 = 1 - 1/z^2 + 1/z^4 - ...
This has essential singularity at z = 0 as that's where the denominator goes to 0.
but book answer is B. How?
 
  • #26
Sushila said:
e^(-1/z^2)=0 has only zero of order 2 at z=0,
It has ##z^{-2}, z^{-4}, z^{-6}, ...##. Looks like a fairly essential singularity to me.
 

1. What is the singularity of e^(-1/z^2)?

The singularity of e^(-1/z^2) is a point in the complex plane where the function becomes undefined, i.e. it approaches infinity or does not have a finite value.

2. Where does the singularity of e^(-1/z^2) occur?

The singularity of e^(-1/z^2) occurs at z = 0, as plugging in 0 for z results in a division by 0 which is undefined.

3. How is the singularity of e^(-1/z^2) related to the concept of complex numbers?

The singularity of e^(-1/z^2) is related to complex numbers because it occurs in the complex plane, where both the real and imaginary parts of z are considered. It is a point where the function behaves differently based on the value of z, which is a key concept in understanding complex numbers.

4. What is the significance of the singularity of e^(-1/z^2) in mathematics and science?

The singularity of e^(-1/z^2) has significance in mathematics and science because it is a point where the function becomes undefined, which can lead to interesting and unexpected results. It also plays a role in the study of complex analysis and the behavior of functions in the complex plane.

5. Can the singularity of e^(-1/z^2) be avoided or removed?

No, the singularity of e^(-1/z^2) cannot be avoided or removed as it is a fundamental property of the function. It is a point where the function behaves differently and removing it would alter the behavior of the function entirely.

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