All Questions
18
questions
1
vote
2
answers
94
views
Proving that inverse of the unit circle parametrization is not continuous. [duplicate]
Statement:
Let us have a continuous and bijective unit-circle parametrization map:
$f: [0, 2\pi) \rightarrow S$
$\phi \mapsto cos(\phi) + i \cdot sin(\phi)$
We prove that $f^{-1}$ is not continuous.
...
- 481
0
votes
1
answer
51
views
Determine derivative wherever the derivative exists of $-i(1-y^2)+(2x-y)(y)$
Is this correct? (Edit: I'm just going to outline the steps and post the rest as an answer.)
$g: \mathbb C \to \mathbb C, g(z) = -i(1-y^2)+(2x-y)(y)$
Step 1. $g$ is differentiable only on $\{y=x\}$.
...
- 13.7k
0
votes
1
answer
40
views
Showing that $f(z) = \sum_{n=1}^{\infty} (n+1)^2 z^n$ is continuous in the open unit disk $|z| <1$
So I figured since its a power series that maybe finding the radius of convergence here would be useful. So as $f(z) = \sum_{n=1}^{\infty} (n+1)^2 z^n$ we can find the radius of convergence as$$R= \...
- 904
1
vote
1
answer
376
views
Complex function with values on the unit circle copied everywhere
If $f:\mathbb{C}\setminus \{0\}\to \mathbb{C}$ is a function such that $f(z)=f(\frac z{|z|})$ and its restriction to unit circle is continous,then
$(1)\lim _{z\to 0} f(z)$ exist.
$(2)f$ is analytic ...
- 2,589
0
votes
1
answer
39
views
Show that the function is continuous which points?
Show that the function is continuous which points.
$$\displaystyle f(z) =
\begin{cases}
\frac{x^2}{x^2+y^2}+2i, & \text{if $z \neq 0$ } \\[2ex]
2i, & \text{if $z=0$ }
\end{cases}$$
...
- 256
0
votes
1
answer
135
views
Continuous square root function on $\mathbb C^*$
There is another thread about this topic but I'm interested if the following reasoning is correct:
Let $f:\mathbb C^*\to\mathbb C^*$ be a continuous function that satisfies $f^2(z)=z$. We know that ...
- 567
3
votes
1
answer
118
views
Where is $k(z)=PV(z-1)^{\frac{1}{2}}PV(z+1)^{\frac{1}{2}}$ Continuous and Differentiable
Consider the function ($PV$ denotes the principal value) $$PV(z-1)^{\frac{1}{2}}PV(z+1)^{\frac{1}{2}}, \ \ \forall z\in\mathbb{C}.$$
Find where $k$ is continuous and differentiable, giving reasons.
...
- 617
0
votes
1
answer
295
views
continuity of $f(z)=(z-i)\log(z^2+1)$
I am trying to determine where $f(z)=(z-i)\log(z^2+1)$ is continuous.
My attempt:
Well, $z-i$ is a complex polynomial, so it is continuous in all of $\mathbb{C}$. Hence, the continuity of $f(z)$ is ...
user557493
1
vote
1
answer
388
views
3 Exercises on uniform convergence of complex sequences and relation to Weierstrass M-Test
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 7.19(a),7.20(a),7.21
What are the errors, if any, in the following proofs? I put $\color{...
- 13.7k
-1
votes
2
answers
684
views
Why is $\int_{[0,1]} \frac{dw}{1-wz}$ is holomorphic in unit disc?
Expanding on the question and answer in: Prove $f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$ is holomorphic in the open unit disk.
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis ...
- 13.7k
1
vote
1
answer
76
views
Continuity of a complex valued function
I have to show that the complex function is continous using ϵ and δ: $ f(z) = |z+2i| $
My attempt:
Assume that $ |z - z_0| < δ $ is true.
I also assume that δ = ϵ. I have to proof the following ...
- 47
2
votes
2
answers
212
views
Proving that this function is continuous on $G\times G$
Let $G\subset \mathbb{C}$ be a non-empty open set and $f$ be a function holomorphic on $G$. Let $g: G\times G\to \mathbb{C}$ be a function defined as $$g(z,w)= \begin{cases}
\frac{f(z)-f(w)}{z-...
- 9,708
0
votes
1
answer
227
views
Limit at set boundary for uniform function
Let $D:=\{z\in \mathbb{C}: \lvert z \rvert < 1\}$ and $B := \{z\in \mathbb{C}: \lvert z \rvert = 1\}$. Let $f:D\to \mathbb{C}$ be a uniformly continuous function. Then $\lim\limits_{z\to z_0, z\in ...
- 9,708
0
votes
1
answer
3k
views
If $f(z)$ is analytic at $z_0$ then it is continuous at $z_0$
My Proof: Since $f(z)$ is analytic at $z_0$ then it is differentiable at $z_0$
1) $f(z_0)$ exists because $f(z)$ is differentiable at $z_0$
2) $\lim_{z\rightarrow z_0}[f(z)-f(z_0)]=\lim_{z\...
- 375
2
votes
0
answers
61
views
Showing that the inverse of $f(t) = e^{it}$ is not continuous by using an open set/pre-image argument
I have the map $f:[0, 2 \pi) \rightarrow S^1, f(t) = e^{it}$. I'm trying to show that the inverse map $g: S^1 \rightarrow [0, 2 \pi), \space \space \space g=f^{-1}$ is not continuous by finding an ...
- 2,909