How do you construct mobius transform easily? is there a certain way to go about it or is it by inspection and experience. for ex. construct one from unit disc to left half plane or to right half plane, or below a certain y=ax+b line or something of the sort.
does the behavior the imaginary part behave in anyway similar to the real part of a holomorphic function. say if the real part if bounded or positive, what can you conclude about the imaginary part.
exactly what are those other functions without z_bar but are still not holomorphic because unless you apply the operator defined by landau it's very hard to tell. Seems like you have to use that operator to make sure. is there anyway to tell on inspection if a function is holomorphic, maybe...
There are a lot of definitions but what is the quickest way to see if a function is holomorphic? apply the cauchy riemann equations seems too slow. I thought if it doesn't have a z_bar in it, then it's automatically holomorphic. so for ex. polynomials are always holomorphic. on the other...
$\frac{d}{dp}(\Vert f\Vert_{p}^{2})\mid_{p=2}=\frac{d}{dp}((\int_{\mathbb{\mathbb{R}}^{n}}\mid f\mid^{p}dx)^{\frac{2}{p}})\ldots=\frac{1}{2}\int_{\mathbb{\mathbb{R}}^{n}}\mid f(x)\mid^{2}ln\left(\frac{\mid f(x)\mid^{2}}{\Vert f\Vert_{2}^{2}}\right)$
this is the derivative evaluated at p=2...
any ideas? if we use any polynomial of order > j like j+1 so f(z) = z^(j+1)
then f(z)(z_bar)^j becomes z(z z_bar)^j = z |z|^2j which is just z since modulus is 1 on boundary of disk of radius 1. so obviously integral of z is 0 since its holomorphic but f(z) is not trivial. am i...
f is a holomorphic polynomial and if
$\oint_{\partial D(0,1)}f(z)\bar{z}^{j}dz=$ 0 for j = 0,1,2,3...
where $\partial D(0,1)$ is the boundary of a disc of radius 1 centered at 0
prove f $\equiv$0
I think I see the confusion, my correction only applies to the 2nd part. the original problem should be
show there is a some constant c, independent of n, s.t. if {Z_j} are complex numbers and sum of |Z_j| from 1 to n >= 1, then there's a subcollection {Z_j_k} of {Z_j} s.t. |sum of Z_j_k| >=...
but obviously the {z1,z2} , {z3} partition does not work for n=3 take z1 , z2 to be a+bi, -a-bi, and z3 to be some other vector perpendicular to them and each length 1/3.
you're saying this is always true for any i? if you have 3 numbers each of which has length 1/3 and if you divide it into z1+z2 and z3 where z1 and z2 point in the opposite directions this wouldn't be true. I don't think its as simple as the triangle inequality