I meant ##C## is a connected component of ##X##. I assumed that if the path ##f## in ##P## is inside the connected subspace ##f([a,b]) \subset C##, then ##P \subset C##; which I think now is insufficient. Is this what you were trying to point out that is deficient? Thanks!
Homework Statement
Theorem: If ##X## is a topological space, each path component of ##X## lies in a component of ##X##. If ##X## is locally path connected, then the components and the path components of ##X## are the same.
I need help locating errors in my proof. Please help.
Homework...
I do have a fair amount of visual/geometric understanding of groups, but when I start solving problems I always wind up relying on my algebraic intuition, i.e. experience with forms of symbolic expression that arise from theorems, definitions, and brute symbolic manipulation. I even came up with...
Homework Statement
Find ##z## in ##z^{1+i}=4##. Is my solution correct
Homework Equations
##\log(z_1 z_2)=\log(z_1)+\log(z_2)## such that ##z_1, z_2\in \{z\in\Bbb{C} : (z=x+iy) \land (x\in\Bbb{R}) \land -\infty \lt y \lt +\infty\}##
##re^{i\theta}=r(\cos\theta + i\sin\theta)##
The Attempt at a...
Just to have closure. I'll put the entire solution here. By letting ##z=e^{i\theta}## and ##dz=ie^{i\theta}d\theta##. we have
\begin{align}
\frac{1}{ie^{i\theta}}\int_{c}\cos^2(\frac{\pi}{6}+2e^{i\theta})ie^{i\theta}d\theta = \frac{1}{i}\int_{c}\frac{\cos^2(\frac{\pi}{6}+2z)}{z}dz
\end{align}...
Got it! Thank you, I think I can handle this from here. Sorry :sorry: if I am taking a while, I am multi-problem solving right now :nb). Exam is tomorrow.:eek:
What is holonomic? Did you meant holomorphic? I googled holonomic and it exists and it means something else.
Yes, I meant formula - sorry. But, the formula is applicable right?
I find it hard to believe they put this on a past exam. It's suppose to be a test of understanding, not puzzles. I'll try my hand at this again later, I'm working on other problems for the moment. Thank you!