All Questions
Tagged with complex-analysis continuity
27
questions
10
votes
4
answers
2k
views
On continuity of roots of a polynomial depending on a real parameter
Problem
Suppose $f^{(t)}(z)=a_0^{(t)}+\dotsb+a_{n-1}^{(t)}z^{n-1}+z^n\in\mathbb C[z]$ for all $t\in\mathbb R$, where $a_0^{(t)},\dotsc,a_{n-1}^{(t)}\colon\mathbb R\to\mathbb C$ are continuous on $t$...
- 9,803
2
votes
1
answer
464
views
Prove the function is continuous, exercise from Conway's "Functions of One Complex Variable I"
For the first proof of Cauchy's integral formula, Conway in his book "Functions of One Complex Variable" (Chapter IV, section 5.4) uses the following claim:
Let $G$ be an open subset of $\mathbb C$ ...
- 3,669
6
votes
1
answer
3k
views
If $f: \mathbb{C} \to \mathbb{C}$ is continuous and analytic off $[-1,1]$ then is entire.
This is a problem from Complex Variable (Conway's book) 2nd ed.
(Section 4.4) 9. Show that if $f: \mathbb{C}\to\mathbb{C}$ is a continuous function such that $f$ is analytic off $[-1,1]$ then $f$ is ...
- 2,505
2
votes
2
answers
212
views
Proving that this function is continuous on $G\times G$
Let $G\subset \mathbb{C}$ be a non-empty open set and $f$ be a function holomorphic on $G$. Let $g: G\times G\to \mathbb{C}$ be a function defined as $$g(z,w)= \begin{cases}
\frac{f(z)-f(w)}{z-...
- 9,708
5
votes
2
answers
8k
views
Proof that a continuous function maps connected sets into connected sets [duplicate]
I'm trying to prove that, if f is a function from C to C, and its domain, D, is connected, then f(D) is also connected. How would I go about doing this?
The definition of conectedness at play is "S ...
- 461
-1
votes
2
answers
684
views
Why is $\int_{[0,1]} \frac{dw}{1-wz}$ is holomorphic in unit disc?
Expanding on the question and answer in: Prove $f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$ is holomorphic in the open unit disk.
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis ...
- 13.7k
3
votes
1
answer
903
views
Analytic continuation of function continuous on boundary
Suppose one has a function $f$ in the disc algebra ie: $f$ is continuous on $|z|\leq1$ and holomorphic in $|z|<1$. I wondered, can $f$ always be extended to a holomorphic function on some region ...
- 873
2
votes
1
answer
100
views
Prove or disprove the uniform continuity of $f: \mathbb{C} \to \mathbb{C}, z \mapsto f(z)=\frac{z^2}{1+|z|}$
I tried using the definition of the continuity to try and find a $\delta$ but I got stuck here
$$ |f(w) - f(z)| = |\frac{w^2 \cdot (1+|z|) - z^2 \cdot (1+|w|)}{(1+|w|)\cdot(1+|z|)}|$$
What can I do to ...
- 129
2
votes
3
answers
2k
views
let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a continuous function and assume $f(z) = f(2z)$, prove that f is constant
$f: \mathbb{C} \rightarrow \mathbb{C}$ be a continuous function and assume that $f(z) = f(2z)$ for all $z \in \mathbb{C}$. Prove that f is constant...
Then we are supposed to use this result to ...
- 95
1
vote
1
answer
676
views
Morera's theorem of entire function
For each fixed $n$, show that $$f_n(z)=\int_1^nt^{z-1}e^{-t}dt$$ is an
entire function of $z$.
From Morera 's theorem:
If a continuous, complex-valued function $f$ in a domain $D$ that
...
- 1,089
6
votes
2
answers
2k
views
No continuous injective functions from $\mathbb{R}^2$ to $\mathbb{R}$
Which of the following statements is true?
$(a)$ There are at most countably many continuous maps from $\mathbb{R}^2$ to $\mathbb{R}$
$(b)$ There are at most finitely many continuous surjective maps ...
- 2,589
5
votes
1
answer
1k
views
$\log$ is continuous
I'm trying to prove that the complex logarithm function is continuous using this theorem, but I'm hitting a snag in part of the proof.
Let $D=\Bbb C\setminus(-\infty,0]$. The claim is that the ...
- 27.7k
4
votes
1
answer
1k
views
Usage of Schwarz Reflection Principle to Study Conformal Equivalence of Annuli
Let $A(1,r) = \{z \in \mathbb{C} : 1 < |z| < r\}$. I would like to prove the standard result that $A(1,r)$ and $A(1,r')$ are conformally equivalent iff $r = r'$. To prove the nontrivial ...
- 2,726
4
votes
2
answers
4k
views
Why $S^1$ does not homeomorphic to $[0, 1)$?
The map $f:[0,1)\to S^1$ given by $f(x)=e^{2\pi ix}$ is a continuous bijection.
However since $S^1$ is compact and $[0,1)$ is not compact, $f$ can not be a homeomorphism.
For any $z\in S^1,$ we have ...
- 18.4k
4
votes
1
answer
557
views
Complex substitution allowed but changes result
It is well known that
$$ I := \int_L \frac{1}{z} ~\text{d}z = 2 \pi i $$
where $L$ is the complex unit circle, parametrized by $\gamma(t) = e^{it}, 0 \leq t \leq 2 \pi$. However, using complex ...
- 566