Thank you for your input.
I totally forgot about complex analysis. I have also taken that class using Palka's Intro to Complex Function Theory. At my university, these are the "most advanced" undergraduate courses.
The algebra class was two-semesters long but as you have already noted...
My curriculum was heavy proofs from the very beginning.
My honors analysis class used Spivak. I also took a masters level analysis class that used baby Rudin. My abstract algebra class used Dummit and Foote. My other classes which include linear algebra, differential equations, number theory...
Its been a long ride. 4 years ago when I started college, I started as a finance major. I excelled in all of my classes but found the material to be a little boring so I changed to economics. I continued to get stellar grades and even now have nothing but A's in all of my economics courses. I...
I figured it out for future reference to anybody.
Use ##sin(z)## to take the infinite strip to ##\mathbb{C}\sim\{w:|\Re(w)|\geq 1## and ##\Im(w)=0\}##. Then rotate this by multiplying by ##i## and finally use ##Arctan(w)## to take it back to the infinite strip.
Homework Statement
Find a conformal mapping of the strip ##D=\{z:|\Re(z)|<\frac{\pi}{2}\}## onto itself that transforms the real interval ##(-\frac{\pi}{2},\frac{\pi}{2})## to the full imaginary axis.The Attempt at a Solution
I tried to map the strip to a unit circle and then map it back to the...
I know that for the tangent unit vector ##t##, normal unit vector ##n##, and binormal unit vector ##b## that ##b=t\times n## and ##n=b\times t##. Is it true that ##t=n\times b##?
**Edit** Ah! Yes it is. Nevermind. I should have known this was true.
This problem comes from the first chapter in the textbook which is an introduction complex analysis. Picard's theorem comes in chapter 4.
Do you know if there is any way to parametrize S without ##x=|z|\cos\theta##, ##y=|z|\sin\theta##?
I can't seem to picture it. In class I was shown that if z=x and Im(z)=0 then e^z was a circle and if Re(z)=0 and Im(z)=y then e^y was a vector that pointed outwards from the origin at an angle of y. Combining these together all I can see is two circles, one inside the other bounding the area...
If ##f(z)=\frac{1}{z}##, ##f(S)## would be a disk with a hole inside it centered at the origin with radius ##\frac{1}{r}## but in this case, the function is the exponential.
I think ##f(z)=e^z## maps z to a circle on the complex plane of radius Re(z) so I'm tempted to say f(S) is a mess of...
If ##0<|z|<r## then we have ##\frac{1}{r}<\frac{1}{|z|}## and ##\frac{1}{|z|}\rightarrow\infty## as ##|z|\rightarrow 0## but ##0<|z|## so we can safely say ##\frac{1}{|z|}<\infty##.
Homework Statement
Determine ##f(S)## where ##f(z)=e^{\frac{1}{z}}## and ##S=\{z:0<|z|<r\}##.
*Edit: The function f is defined as ##f:\mathbb{C}\rightarrow\mathbb{C}##.
The Attempt at a Solution
I am a little confused as to what this problem is asking me to do. What I did was:
Let...
Ah! Then since ##z_{0}## is contained in ##f(\mathbb{C})##, this is a contradiction because ##G## does not contain its boundary. If it did then ##G## would also have to be closed and the only sets which are both open and closed in ##\mathbb{C}## are ##\emptyset## and ##\mathbb{C}## but since...
By Bolzano-Weierstrauss, there exists a convergent subsequence of ##z_{n}##,##z_{n_{k}}## which converges to some ##z\in\mathbb{C}##. Then since ##f## is continuous, ##f(z_{n_{k}})\rightarrow f(z)=z_{0}##.
I am not sure how to proceed. I keep thinking that the goal is to derive a contradiction...
Homework Statement
Let a continuous function ##f:\mathbb{C}\rightarrow\mathbb{C}## satisfy ##|f(\mathbb{C})|\rightarrow\infty## as ##|z|\rightarrow\infty## and let ##f(\mathbb{C})## be an open set. Then ##f(\mathbb{C})=\mathbb{C}##.
The Attempt at a Solution
Suppose for contradiction that...