# Tag Info

### How is $|z|^2$ not analytic?

Let me summarize the points already made in comments: The definition of "analytic at $p$" is "differentiable in some neighborhood of $p$". I.e., it is not equivalent to "...
• 43.7k
1 vote

• 21.8k
1 vote

• 2,212
1 vote

### Complex Integral and Residue involving multiple Branch Points inside the contour, without deforming it

The strategy will depend a lot on $f$. I'll assume for simplicity that $f$ is meromorphic. Technically your integrand is ill defined since the square root is double valued: you need to specify the ...
• 2,948
1 vote

### Show that $f(\frac{1}{z})$ have a essential singularity at $0$.

Yes, that's enough. If $z=0$ is removable for $f(1/z)$, then $f(1/z)$ is bounded around $0$, therefore $f$ is bounded around $\infty$, hence bounded over the whole $\mathbb C$. To be a little bit more ...
• 15.2k
1 vote

### Show that $f(\frac{1}{z})$ have a essential singularity at $0$.

$f(\frac 1 z) \to f(0)$ as $|z|\to \infty$. If $f(\frac 1 z)$ has a removable singularity at $0$, then it extends to a bounded entire function. Hence, it would be constant, by Liouville's Theorem.
• 37.1k
1 vote

### Show that $f(\frac{1}{z})$ have a essential singularity at $0$.

Assume that $f(1/z)$ has removable singularity at $z=0$. Then we know that this can happen only if $f(1/z)$ is bounded near $z=0$. But this means $f(z)$ is a bounded entire function, hence is constant ...
• 333

### Show that $f(\frac{1}{z})$ have a essential singularity at $0$.

Since $f(z)$ is entire, we may write it as $f(z)=\sum_{k=0}^\infty a_kz^k$. Since $f$ is non-constant, there is some $a_k\neq 0$ where $k>0$. We will show infinitely many of the $a_k$ are non-zero. ...
• 2,285
1 vote
Accepted

### Computing Singularities of a function

It doesn't matter if the limit of the mentioned expression tends to zero. The only thing that matters for a function like that is the existense of the limit.
1 vote

### Examples of non-elliptic Doubly Periodic Functions

Complex analysis is a very rigid subject: Given the values of functions on a countable complex set with a finite accumulation point, the function is fixed inside a circle of convergency around this ...
• 2,212

### Prove that, except for the identity function, a holomorphic map of the open unit disk into itself has at most one fixed point in the disk.

Let $z_{1},z_{2}\in B(0,1)$ be distinct ,$f\in H({B(0,1)}) ,f(z_1)=z_1$, $f(z_2)=z_2$. Def:$$\phi_{a}(z)=\frac{a-z}{1-\bar{a}z}$$ Notice that:$$\phi_{z_1}(f(z_2))=\phi_{z_1}(z_2)$$ Using Schwarz-...
1 vote

### Examples of non-elliptic Doubly Periodic Functions

Elliptic functions come from inverting elliptic integrals. In particular, when $0<k<1$ and $$w=\int_0^z{\mathrm dt\over\sqrt{1-t^2}\sqrt{1-k^2t^2}}.$$ The inverse $z=\operatorname{sn}(w)$, ...
• 6,788

### Prove that, except for the identity function, a holomorphic map of the open unit disk into itself has at most one fixed point in the disk.

Hints: Let $c,d$ be distinct fixed points of $f$. Define $g=\phi^{-1}\circ f\circ \phi$ where $\phi (z)=\frac {z+c} {1+\overline c z}$. Then $g(0)=\phi^{-1}c=0$. Since $\phi$ maps the unit disk to ...
• 37.1k
1 vote
Accepted

### exponential map property

The answer is yes. You can even take some larger domain for $f$: any interval $I$ of length $L<\pi$, to make sure that $$\frac{|f(s)-f(t)|}2\le a:=\frac{|e^{iL}-1|}2<1\quad(\forall s,t\in I).$$ ...
• 35.3k

• 333

• 27.5k
1 vote

### Solve complex equation $(z-i)^4=(1+2i)^8.$

Since $(1+2i)^2= -3 + 4i$, the equation can be written as $(z-i)^4= (-3+4i)^4$. Note that for $a\in\mathbb{C}$, the solutions of $z^4= a^4$ are $z\in\{\pm a, \pm ai\}$. Hence the roots of the above ...
• 6,515
Accepted

### Confusion in proof of $H^1(X,\mathcal{O})=0$ where $X$ is an open disk

Your shortened proof is true, but it's circular until you know the cohomology of the disk vanishes. The definition of Cech cohomology says we have to take a direct limit over all the opens; Leray's ...
• 30.3k
Accepted

### Calculate $\int_{-\infty}^{\infty}\frac{e^{itz}}{(z+i)^2}\,\mathrm{d}z$

The correct answer is $$\int_{-\infty}^{\infty} \frac{e^{itz}}{(z+i)^2} \, \mathrm{d}z = \begin{cases} 0, & t \geq 0, \\[0.25em] 2\pi t e^t, & t < 0. \end{cases}$$ So, what have gone ...
• 168k

### Calculate $\int_{-\infty}^{\infty}\frac{e^{itz}}{(z+i)^2}\,\mathrm{d}z$

Trying to calculate $$\int_{-\infty}^{\infty}\frac{e^{itz}}{(z+i)^2}\,\mathrm{d}z$$ ...for $t > 0$, I use the lower semicircle that encapsulates the singularity at $z = -i$. ...$= 2\pi te^t$ ......
• 556

• 56
Accepted

### Non-constant holomorphic function definied on open disk $D(0,1)$

Take $r_1,r_2\in[0,1]$ such that $r_1<r_2$. We know that $h(r_2)$ is attained at the circle of radius $r_2$. If $h(r_1)=h(r_2)$ then there would exist some $x_1$ in the circle of radius $r_1$, and ...
• 4,391
1 vote

### Non-constant holomorphic function definied on open disk $D(0,1)$

The maximum of $|f|$ on $\overline {D(0,r_2)}$ is attained at some point on its boundary. If $h(r_1)=h(r_2)$ the maximum would also be attined at some point with $|z|=r_1$. By Maximum Modulus ...
• 37.1k
Accepted

### $f$ with an essential singularity at $0$ intersects $1/z^4$ in every neighborhood of $0$.

$z$ is an element of the image of $g$, it may not even be near $0$. Nonetheless, the proof is still not very difficult. As $\dfrac{1}{z^4}$ has a pole at $0$, $f(z)-\dfrac{1}{z^4}$ has an essential ...
• 4,391
1 vote
Accepted

### Uniform convergence of $\sum_{n=1}^{\infty} \frac{\exp(-nz)}{\sqrt{n(n+1)}}$

Another way is to use Cauchy's criterion for uniform convergence, and show it fails. Since the series $\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n(n+1)}}$ diverges, there is some $\epsilon>0$ such ...
• 40.2k

### Finding linear imaginary and real factors of cubic P(z).

$$P(z)=z^3+az^2+3z+9$$ Let zeroes be represented as $p,iq,-iq$ where $p,q \in R$, because complex roots occur in conjugates, and here they are purely imaginary. Sum of roots $=-a=p+iq-iq=p$ $p$ ...
• 1,135
1 vote

### Uniform convergence of $\sum_{n=1}^{\infty} \frac{\exp(-nz)}{\sqrt{n(n+1)}}$

Your justification is sufficient. Actually, we have the following result: Suppose $z_0$ is a limit point of $E\subset\mathbb C$, the series $\sum\limits_{n=1}^{\infty}u_n(z)$ is uniformly convergent ...
• 7,265