Expert 4x

1) I need to find the Fourier series of f (x) <= x + pi-pi x <= pi, beeing f (x) periodic. when I try to find the coefficients h bm y, I bm = 0 for all m. I must be an error on the full but I can not find it!

2) For f (z) = ((zi) / (z + i)) ^ (1 / 2)
a) Find the dominium where f (z) is analytic, is C-(o,-i)?, I mean the plan every coplas except the imaginary axis between zero and-i.
b) Rate of convergence of Taylor series to 2i, and that's all. I could find, but the products was complicated, I want to know if there is a trick to avoid these derivatives, such as working on a simple series.
3) if you are a mathematical genius: If the harmonic function u (x, y) has a null value in the x ^ 2 + y ^ 2 = 1 unit circle, prove that is void in all its dominium

Twopointers please refrain. Do not make me lose my time, as you do. If you can not help, move to where you can.

Part 1

To do this, in a canonical form that the period is equal to 2 feet. (I guess that the period you want to say is-pi <x pi ").

The first thing is to split your integral aˆ« (x + s) cos (x) dx and aˆ« (x + s) sin (x) dx by distinct integral aˆ« x * cos (x) dx + aˆ« pi * cos ( x) dx and aˆ« x * sin (x) dx + aˆ« sin * ft (x) dx.

You should get:

a0 = p

cos (0) = 1, aˆ« x time = 0 but aˆ« term pi = pi

h = 0
BM = 2 * ((-1 ^ (m +1)) / m)

aˆ« pi * cos (x) dx and aˆ« sin * ft (x) dx is equal to 0 because it is a constant times cos or sin integrated over a whole number of periods. aˆ« x * cos (x) dx and aˆ« x * sin (x) dx are exactly like the function of the identity periodicals.

Are you sure you do intergating during the period <-ft x <pi.

Part 2

Do not have an answer for this moment. If I get it, I'll put up as an edit later.

PS Hope displays ok integral sign, otherwise it may seem a bit odd. Hope this helps.

Edition -

Sorry, can not help you with parts 2 and 3. The denominatorof f (z) aˆš (z + i) clearly has a root at z = (0,-i), but other than that I am not sure of the question. Hope the advice of Part 1 have been a little help.

1 = 1q has 2P4> me.

For part 1, you seem to have a very simple function that does not need a Fourier series. Are you sure it is properly written in the question?
For part 2a, the function is defined for C - (0,-i). This means that the whole complex plane, except the single point (0,-i) not part of the imaginary plane between the present and the origin. (You should not have said that part of the imaginary axis from 0 to - 1, you do not need me because you've already said that this axis.)
I think I can also give you a little help with the rest of Part 2.
You should find that the first derivative can be rearranged to
i * f (z) / (z + i) ^ 2. (This may not be the case as I have done quite quickly.) Yes, making it easier to find the second derivative. The same thing could happen again, but I have not been so far.