Let ##\gamma## be a closed curve in ##\mathbb{C}##. If ##\gamma## doesn't contain any point from [2,5] in its interior, then ##\int_{\gamma}f=0## since f is holomorphic away from [2,5]. Suppose that ##\gamma## contains [2,5] in its interior. Let R be a rectangle oriented with the coordinate...
For your solution when k=1, why do you have [-1,\infty)?
If t = -1, you are dividing by zero.
What is wrong with the (-\infty, -1)?
(-2)^2-4+1>0 Is this true for the entire set?
I have this proof for finite points but how would I modify it for infinite many points between [2,5]?
Assume q(z) is any function that is holomorphic on a disc U except at a finite number of points \xi_1,\ldots, \xi_n\in U, and assume \lim_{z\to\xi_j}(z-\xi_j)q(z)=0 for 1\leq j\leq n. Let...
So I am trying to use Morera's Theorem:
Let U be an open set in C and let f be continuous on U. Assume that the integral of f along the boundary of every closed rectangle in U is 0. Then f is holomorphic.
So let U = \mathbb{C} - [2,5] Let R be rectangles in U which are parallel to the...
If you set I = integral and multiplied by the same integral, you would have I^2. When you solve that you get 2pi but then you take the square root.
However, isn't is supposed to be equal to \frac{1}{\sqrt{2\pi}
Then why did you ask if you knew that part?
What you have as a proof doesn't look like you used the polar identity. I am not really sure what you have done to claim <T(x),T(y)>=<x,y>
If you use the polar identity, your next step should be
\frac{1}{4}\left(||T(x+y)||^2-||T(x-y)||^2\right)
=\frac{1}{4}\left(||x+y||^2-||x-y||^2\right)=\langle x,y\rangle
It seems ok but could be written better.
Like you said m is a product of primes so lets write m as m=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_t^{\alpha_t}. Then m^n = \left(p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_t^{\alpha_t}\right)^n = p_1^{n\alpha_1}p_2^{n\alpha_2}\ldots p_t^{n\alpha_t}. Since...
If the pole is at z_0, then
f=\frac{g}{(z-z_0)}
Let S be the set of the poles of f in U and let h be analytic in U with singularities at the points in S. Let the order of the singularity be the order of the pole in f. Then fh has only removable singularities in U.
fh=g\Rightarrow f=\frac{g}{h}
If f is meromorphic on U with only a finite number of poles, then f=\frac{g}{h} where g and h are analytic on U.
We say f is meromorphic, then f is defined on U except at discrete set of points S which are poles. If z_0 is such a point, then there exist m in integers such that (z-z_0)^mf(z)...