Hi all, i have a couple of questions on my complex analysis course that i cannot do and was wondering if you could help at all.
QUESTION 1
a) Suppose that the functions f(z) = u(x,y) + iv(x,y) and g(x) = v(x,y) + iu(x,y) are analytic in some domain D. Show that both u and v are constant functions
b) Let f be a holomorphic function on the punctured disk D'(0,R) = {z in complex plane : 0 < |z| < R}, where R > 0 is fixed. Write down the formulae for c(subscript)n in the Laurent expansion
f(z) = sum from n = minus infinity to infinity c(subscript)n . z^n (by this i mean cn multiplied by z^n)
Using these formulae, prove that if f is bounded on D'(0,R), it has a removable singularity at 0.
c) FInd the maximal radius R > 0 for which the function f(z) = (sin z)^-1 is holomorphic in D'(0,R), and find the principal part of its Laurent expansion about z(subscript)0 = 0 (centred at 0)
(the z(subscript)0 refers to: sum from k = minus infinity to infinity c(subscript)k(z - z(subscript)0)^k the Laurent expansion)
QUESTION 2
Find the Laurent exansions of the function f(z) = z / (z^2 - 1) valid for:
i) 0 < |z - 1| < 2
ii) |z + 1| > 2
iii) |z| > 1
Thank you
Doug
The theorem says that we can write,
where .
But we do not need this to prove that if is bounded then it has a removable singularity.
If then .
But thus, .
If then is defined as long as .c) FInd the maximal radius R > 0 for which the function f(z) = (sin z)^-1 is holomorphic in D'(0,R), and find the principal part of its Laurent expansion about z(subscript)0 = 0 (centred at 0)
(the z(subscript)0 refers to: sum from k = minus infinity to infinity c(subscript)k(z - z(subscript)0)^k the Laurent expansion)
Thus, this radius is .
But the formula will be used to prove that
"If then is a removable singularity."
In order to prove this we have to show that all 's are 0 (this is the definition of removable singularity).
( is a circle of radius r centered at )
We know that f is bounded, ,
Now we can use the formula.
And goes to 0 for smaller circles.
Thank you both very much for your quick replies and great responses. They have helped alot.
For the question 2, about the Laurent expansions I have:
z/(z^2 - 1)
by partial fractions this becomes:
1/2(z - 1) + 1/2(z + 1)
so for:
0 < |z - 1| < 2
1/2(z + 1) = 1/2(2 + z - 1) = 1/4(1 + (z - 1)/2)
= 1/4 x (sum from n to infinity) (-1)^n . [|z - 1|/2]^n
this proves |z - 1| < 2
not too sure how to prove that |z - 1| > 0 though.
for:
|z + 1| > 2 i have:
1/2(z - 1) = 1/2(-2 + z + 1) = 1/2 . 1/(z + 1) . 1/(1 - 2/(z + 1))
= 1/2 . 1/(z + 1) . (sum from n to infinity) [ 2/|z + 1|]^n
which proves |z + 1| > 2
That is what i have so far, i am not sure how to do the other half of the partial differential equation in both cases. To prove it totally, also a little unsure on the third part.
Thank you once again
Doug
That's correct. You have found the series expansion of 1/(2(z+1)) (in powers of z-1) which converges when |z - 1| < 2. But don't forget to include the other partial fraction 1/(2(z-1)). This is a negative power of z-1, so it is only defined when z≠1, or in other words when |z - 1| > 0.
Again, the calculation is correct, but it's not appropriate to say that it "proves" |z + 1| > 2. What you should asy is that this series converges when |z + 1| > 2. (Here too, don't forget to include the other part of the partial fraction expression.)
For the third part of the problem, in the region |z| > 1, you need to get series expansions of both parts of the partial fraction expression:
I think your approach was better than mine because it is like the poster asked.
However, the proof of the theorem (which is quoted) does not need to proven with approximating the integral. Remember a point is a removable singularity if there exists a function in the punctured neighborhood of such that for all and is analytic at . Now to prove this define the function for and . Then clearly is analytic on the punctured neighrhood and continous at . It follows that is analytic on the entire disk. Finally, define for and . Then is analytic on the entire disk and for all . Thus, has removable singularity.