Recent content by jsi

Homework Statement Compute the integral: ∫ x2/(x4-4x2+5) Homework Equations Uses Residue theorem. The Attempt at a Solution So I found the zeroes of x4-4x2+5 to be 2+i and 2-i, and therefore the one that is of relevance is 2+i since it is in the upper half-plane. Then I used...

I ended up getting it I think: I set g(z) = e(f(z)) which is an entire function and then |g(z)| = e^(Re(f(z))) and since Re(f(z)) is bounded by n because Re(f(z)) ≤ n as the problem stated, |g(z)| ≤ e^(n) which is a constant so g(z) is constant and since g(z) = e^(f(z)) it follows that ln(g(z))...

applying cauchy's theorem to g(f(z)) would show that it is entire because all points are in the radius of |f(z)| less than or equal to n?

ok that's super easy then! If I were to do it with g(z) = e^(f(z)) would that also be entire so then it would be bounded and therefore constant and would follow the same way for f(z) to be constant? Or is that a little different? And how did the real part of f(z) being less than or equal to n...

If g(z)=1/((n+1)-f(z)) like you stated, then f(z) must be constant as well since it's in a function that is constant?

Yes, so then |g(z)| must be less than or equal to some constant?

Yes, that's bounded, but I'm really not sure how that relates to Liouville's Theorem =/ also, then if I did e^(f(z)), it would give me the imaginary part of the f(z) but since the problem is asking to consider the real part of f(z), e^(Re(f(z))) would just be e^(some real function)? Again...

so e^(f(z)) = cos(f(z)) + isin(f(z)) or since it's looking for Re(f(z)) should it be just cos(f(z)) since that's the real part of Euler's formula?

I haven't made any progress on this and I'm still pretty confused...

So am I going to want to show g(z) = (exp(f(z)) - exp(f(0))) / z and apply Liouville's Thm which would then show exp(f(z)) = exp(f(0)) which shows f(z) = f(0) then f is constant?

Homework Statement Given: f is an entire function, Re f(z) ≤ n for all z. Show f is constant. Homework Equations The Attempt at a Solution So I thought I'd use Liouville's Theorem which states that, if f(z) is entire and there is a constant m such that |f(z)| ≤ m for all z...

great, thank you!

I think I got it; f'(0) = 0 implies that a1 = 0 and f''(0) = 0 implies that a2 = 0, so f to the nth derivative of 0 = 0 implies that an = 0, so all an must be 0?

oh ok, so if say, since f(z) = 0, d/dz = 0, then taking the derivative of every term of the series yields d/dz (an(z - z0)n) = (nan(z - z0)n-1) = (nan(z - z0)n)/(z - z0)1 and then that means (z - z0) can't = 0 so the ai must be 0? Is that right, or sort of right, or totally wrong?

Homework Statement If f(z) = \sum an(z-z0)n has radius of convergence R > 0 and if f(z) = 0 for all z, |z - z0| < r ≤ R, show that a0 = a1 = ... = 0. Homework Equations The Attempt at a Solution I know it is a power series and because R is positive I know it converges. And if...