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

[jaus tail]

Homework Statement

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?

 

[Dick]

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.

 

[fresh_42]

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.

 

[Dick]

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.

 

[fresh_42]

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.

 

[jaus tail]

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.

 

[Dick]

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 $f(i/10)$?

 

[fresh_42]

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 $t \in \mathbb{R}\, , \,t \to 0\,?$

 

[jaus tail]

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?

 

[Dick]

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.

 

[jaus tail]

But book answer is B.

 

[fresh_42]

But book answer is B.

Again, is it $f\, : \,\mathbb{R} \longrightarrow \mathbb{R}$ or is it $f\, : \,\mathbb{C} \longrightarrow \mathbb{C}$ with $f(z)=\exp (-\frac{1}{z^2})\,?$

 

[jaus tail]

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?

 

[jaus tail]

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.

 

[fresh_42]

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!

 

[Dick]

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.

 

[jaus tail]

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.

 

[fresh_42]

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.

 

[jaus tail]

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?

 

[Dick]

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.

 

[jaus tail]

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.

 

[Dick]

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.

 

[jaus tail]

Wow thanks. I hope this question comes in exam now. Direct 10 marks, ten such questions. :D

 

[Sushila]

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

 

[Sushila]

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

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?

 

[haruspex]

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.