Complex Analysis pole-order problem - did I do it correctly?

1. Nov 13, 2013

Nikitin

1. The problem statement, all variables and given/known data
Determine the location of the singularities, including those at infinity. For poles also sate the order.
$$f(z) = \frac{1}{(z+2i)^2}-\frac{z}{z-i}+\frac{z+i}{(z-i)^2}$$

2. Relevant equations
Theorem: If a function $f(z)$ has a zero of nth order at $z_0$, then the function $h(z)/f(z)$ has a pole of order $n$ at $z_0$, provided h(z) is analytic at $z_0$.

3. The attempt at a solution
Singularities exist at z=-2i and z=i, obviously.

To find the pole order, I split up f(z) into 3 functions a(z),b(z) and c(z). One function for each fraction in f(z). Then I use the theorem above to find that a(z) has a pole at -2i of order 2, b(z) has a pole of order 1 at z=i, and c(z) has a pole of order 2 at z=i.

So I figured f(z) has a pole of order 2 at -2i, and one of order 2 at i. I mean, if I were to add the expansions of functions a, b and c around z=-2i, I would get a total laurent series of order 2. If I did the same around z=i, I would get a laurent series of order 2.

Did I do it correctly? I'm a bit shaky on this stuff.

Oh, and what are the quickest ways to know that a function $h(z)$ is analytic? Cauchy Riemann conditions? Memorization?

Last edited: Nov 13, 2013
2. Nov 13, 2013

Dick

Yes, that sounds right. You recognize functions to be analytic mostly because they are combinations (sums, products, quotients, compositions etc) of simpler functions that you already know are analytic.

3. Nov 13, 2013

HallsofIvy

Staff Emeritus
You could also have added the fractions. You will obviously get some polynomial over $(z+ 2i)^2(z- i)^2$. Since, as you say "If a function f(z) has a zero of nth order at z0 , then the function h(z)/f(z) has a pole of order n at z0 , provided h(z) is analytic at z0. f will have poles of order 2 at both -2i and i.

However, your statement is not quite correct. If h also has a zero at z0 that will decrease the order of the pole- if the order of the zero in the numerator equals the order of the zero in the denominator, that can even change the pole to a "removable discontinuity".

Here, however, when you add the fractions, your numerator is i(z- 1) which has a 0 at z= 1 but no zero at i or -2i.

4. Nov 13, 2013

Nikitin

Hallsofivy: yes, I forgot if h(z_0) is zero the theorem doesn't hold. thanks. But why and how is the order of the pole decreased by the presence of that zero? Also,

How do you know the rational function over $(z+ 2i)^2$ and $(z- i)^2$ are analytic at z=-2i and z=i, respectively? Is it because a rational function can always be regarded as a series expansion (of only 1 term), always?

Dick: Hmm.. That's a good argument. So the expansions of non-analytic functions are combinations of functions which are not analytic? Or non-existent even???Is this the fastest way to find out if a function is analytic - try to find its expansion in a book ?

Last edited: Nov 13, 2013
5. Nov 13, 2013

Dick

What I mean is that I know sin(3*z^2) is analytic because I know f(z)=z is analytic, so z^2 is analytic because it's the product f(z)f(z) of two analytic functions. Same for 3z^2 because the constant 3 is analytic, so h(z)=3z^2 is analytic. And finally g(z)=sin(z) is analytic, so g(h(z))=sin(3z^2) is analytic because it's the composition of analytic functions. I don't have to try to find its expansion.

What kind of functions are you talking about?

6. Nov 13, 2013

Nikitin

So multiplications of analytic functions with each other produce an analytic function? How about dividing? 1/z is not an analytic function on the whole complex plane, yet both 1 and z are.

But how do you know sin(f(z)) or e^f(z) is analytic just because f(z) is? Is it because sin and exp are "analytic" operations or something?

I would appreciate if you could link me to somewhere where this "an analytic function is a composition of other analytic ones" stuff is explained?

7. Nov 13, 2013

Dick

It's just like the 1/z example. If f(z) and g(z) are analytic then f(z)/g(z) is analytic except where g(z) is zero. And all you need to know about compositions is that f(g(z)) is also analytic. So if f(z)=e^z (which I know is an analytic function), then f(g(z))=e^g(z) is also analytic if g(z) is.

8. Nov 13, 2013

Nikitin

So you're telling me an arbitrary analytic function f(g) will ALWAYS produce an analytic function when g(z) is also analytic? Do you have some kind of intuitive proof of this?

Could you provide examples of functions which are non-analytic?

9. Nov 13, 2013

Dick

It's pretty easy to understand. If f(z) is analytic (say at z=0) then you can write it as a convergent power series f(z)=a0+a1*z+a2*z^2+..., then f(g(z))=a0+a1*g(z)+a2*g(z)^2+... and all of the terms in that series are analytic, since products and sums are analytic. The most famous nonanalytic function is complex conjugation. If z=a+bi then f(z)=a-bi is NOT analytic.