Thread: a few complex calculus questions

1. a few complex calculus questions

http://www.math.ualberta.ca/~runde/files/ass411-2.pdf
I believe I got #1, 4, 5, 6i)
for 2, i parametrized f(t) = t+it , t belongs to [0,1]
the its
integral from 0 to 1 of (2it)e^(2i(t^2)) dt
from there I don't know what to do
and I don't know if that is right either

3 I think is partially done for me in my notes, I'll see if I can do it
but hints are appreciated

4 I used the unit circle at (1,1) showed the integral wasn't 0 and then said the function wasn't analytical since the integral wasn't 0 on a closed curve <- can someone confirm this is right?

6ii) I'm not sure what to do there especially since my proof for i) isn't that great

2. Originally Posted by jbpellerin
for 2, i parametrized f(t) = t+it , t belongs to [0,1]
the its
integral from 0 to 1 of (2it)e^(2i(t^2)) dt
from there I don't know what to do
and I don't know if that is right either
Hint: The function you are integrating has a primitive.

3 I think is partially done for me in my notes, I'll see if I can do it
but hints are appreciated
Let $\displaystyle X$ be the set of all points in $\displaystyle D$ which can be polygonally connected and $\displaystyle Y$ be the set of all points in $\displaystyle D$ which cannot be polygonally connected. Prove that $\displaystyle X$ and $\displaystyle Y$ are open sets. Now since $\displaystyle X\cap Y = \emptyset$ and $\displaystyle X\cap Y = D$ it follows that $\displaystyle X= D$ or $\displaystyle Y=D$ by definition of connectdness of $\displaystyle D$.

Note: This argument can be used to show that there is a polygonal path consisting of only vertical and horizontal segments.

4 I used the unit circle at (1,1) showed the integral wasn't 0 and then said the function wasn't analytical since the integral wasn't 0 on a closed curve <- can someone confirm this is right?
If there is $\displaystyle f:\mathbb{C}\to \mathbb{C}$ with $\displaystyle f'(z) = \bar z$ then $\displaystyle f$ is twice differenciable too (in fact it is infinitely differenciable) and this would mean that the mapping $\displaystyle z\mapsto \bar z$ is differenciable - but that is false - it is not differenciable at $\displaystyle 0$.

3. Originally Posted by ThePerfectHacker
Hint: The function you are integrating has a primitive.
so is the integral simply $\displaystyle e^{(2i)}-1$

cool I just learned how to write some nice looking math on this site haha