New answers tagged

0

You are making the same mistake in both exercises: in principle, $M(z) = \dfrac {az + b} {cz + d}$. When you analyze $M(-1) = 0$ you should obtain that $\dfrac {-a + b} {-c + d} = 0$, so $b = a$, therefore your transformation looks like $M(z) = \dfrac {az + a} {cz + d}$, so after solving the first equation the number of unknowns goes from $4$ to $3$. What ...


-1

This way is correct, but there is a method that is more fast. If you want the Moebius trasform that maps the points $z_1,z_2, z_3$ in the points $w_1,w_2,w_3$ you can solve the equation (in $w$) $\dfrac{(z-z_2)(z_1-z_3)}{(z_1-z_2)(z-z_3)}=\dfrac{(w-w_2)(w_1-w_3)}{(w_1-w_2)(w-w_3)}$ and obtain your Moebius trasform . ps:When you multipling for the conjugate ...


1

If you really want to do this using Wirtinger derivatives, $$ u = \operatorname{Re}(f) = \frac12(f + \bar f), $$ so $$ \Delta u = 2 \frac{\partial^2 f}{\partial z \partial \bar z} + 2 \frac{\partial^2 \bar{f}}{\partial z \partial \bar z} = 0 $$ since $\dfrac{\partial f}{\partial \bar z} = 0$ and $\dfrac{\partial \bar f}{\partial z} = ...


0

We can write in the form $\;z=e^{it}\;,\;\;t\in\Bbb R\;$ , so $$z^n=1\iff e^{int}=1\iff nt=2k\pi\;\;\;k\in\Bbb Z\iff t=\frac kn2\pi$$


1

You do some mistakes. Note that $$\frac{\partial}{\partial x} (g \circ f) = (\frac{\partial g}{\partial x} \circ f)(\frac{\partial f}{\partial x})+(\frac{\partial g}{\partial y} \circ f)(\frac{\partial f}{\partial x})$$ and $$\frac{\partial}{\partial y} (g \circ f) = (\frac{\partial g}{\partial x} \circ f)(\frac{\partial f}{\partial y})+(\frac{\partial ...


3

There is a proof due to Nelson that a bounded harmonic function is constant using the mean value properties of harmonic functions.


1

Not sure which Cauchy's theorem you are referring to, but if you allow the weaker one( i.e. cauchy's theorem for a circle contour. You only need Lebniz rule to prove this) , you can prove Liouville's theorem using the independence of path and Lebniz rule. Check out "Conway - Functions of one complex variable"


1

Observe that $$ f'(z)=zf(z)\,\,\Rightarrow\,\,\mathrm{e}^{-z^2/2}\big(f'(z)-zf(z)\big)=0 \,\,\Rightarrow\,\,\big(\mathrm{e}^{-z^2/2}f(z)\big)'=0, $$ and thus $\mathrm{e}^{-z^2/2}f(z)$ is constant. In particular $$ \mathrm{e}^{-z^2/2}f(z)=\mathrm{e}^{-0^2/2}f(0)=1, $$ and thus $$ f(z)=\mathrm{e}^{z^2/2}. $$


1

$|e^{a+bi}|=|e^a|\cdot|e^{bi}|=|e^a|=e^a$. On the other hand, $e^{|a+bi|}=e^{\sqrt{a^2+b^2}}$, and as you noted, $a\leq\sqrt{a^2+b^2}$ for every $a\in\mathbb{R}$. Therefore, $|e^z|\leq e^{|z|}$ for every $z\in\mathbb{C}$, as $e^x$ is monotonically increasing for $x\in\mathbb{R}$ and $a,\sqrt{a^2+b^2}\in\mathbb{R}$ with $a\leq\sqrt{a^2+b^2}$.


5

$$\left|e^z\right|=\left|e^{x+iy}\right|=\left|e^x\right|\left|e^{iy}\right|=e^x\leq e^{|z|}$$ since $x\leq|z|$ and the exponential function is monotonic increasing.


1

Recall that we have $$ e^z = e^{\Re z}e^{i\Im z} $$ As $|e^{i\Im z}| = 1$, we have $|e^z| = e^{\Re z}$. As $\Re z \le |z|$ and the exponential is monotone on $\mathbf R$, we have $$ |e^z| = e^{\Re z} \le e^{|z|}. $$


2

It is correct. For all $z \in \Bbb C$, $\text{Re} \, z \le |z|$ and therefore $$ |e^z| = e^{\text{Re} \, z} \le e^{|z|} $$


2

Consider the Taylor expansion, and apply the regular absolute value inequality to it for each finite term, then take the limit.


1

In the following I will denote by $B_r(a)$ the open disk $\{ z \in \Bbb{C} : |z-a|<r\}$. Let $K \subset G$ be any compact subset. In particular, the distance $d=d(K,\Bbb{C} \setminus G)$ is not zero. This means that you can consider the open cover of $K$ inside $G$ $$K \subset \bigcup_{a \in K} B_d(a) \subset G$$ Since $K$ is compact, you are allowed to ...


1

I don't know how you got $\phi^2(z)=e^{i\theta}\bar c+z$, but: $$\phi\circ \phi(z)=\phi(e^{i\theta}\bar z+c)=e^{i\theta}\overline{e^{i\theta}\bar z+c}+c=e^{i\theta}(e^{-i\theta}z+\bar c)+c=z+e^{i\theta}\bar c+c$$


1

You must have $\mid F_1(i)-F_1(0)\mid =\mid i-0\mid =1$, this implies $F_1(i)=e^{ic}$. If you write $\mid F_1(i)-F_1(1)\mid^2 =\mid i-1\mid^2 =2=\mid e^{ic}-1\mid^2= (cos(c)-1)^2+sin^2(c)=2(1-cos(c))=2$. It implies $c=\pi/2$ or $c=-\pi/2$. So $F_1(i)=i$ or $F_1(i)=-i$.


0

$$\frac{\overline z^2+4}{z^2+4}\,|z+2i|^2=\frac{(\overline z-2i)(\overline z+2i)}{(z+2i)(z-2i)}(\overline z-2i)(z+2i)=\frac{(\overline z-2i)^2(\overline z+2i)}{z-2i}\xrightarrow[z\to+-2i]{}?$$ You can now check which limit ($\;z\to2i\;\;or\;\;z\to-2i\;$) exists and which one doesn't.


1

Without Laurent series and assuming $\;f(z)\;$ isn't zero (because then it is trivially true). By the given information there exists $\;M\in\Bbb R^+\;$ such that $\;|f(z)|>1\;\;\;\forall\,z\in\Bbb C\;\;\text{with}\;\;|z|>M\;$ . It must be that $\;f(z)\;$ has a finite number of zeros $\;z_1,...,z_n\;$, otherwise its set of zeros, which is in ...


5

Suppose $f$ has Taylor series $$ f(z)=\sum_{n=0}^{\infty}a_nz^n \quad\text{and }\quad g(z) =f(1/z) =\sum_{n=0}^{\infty}\frac{a_n}{z^n} $$ If $f$ is not polynomial, then $0$ is an essential singularity of $g$ ($\infty$ is an essential singularity of $f$). By Casorati-Weierstrass theorem, for any $A\in \Bbb{C}$, there is a sequence $z_n\to0$ such that ...


1

Not like that. Instead: $$ 0\le\left|\frac{xy^3-0}{x+iy-0}\right|=\frac{|xy^3|}{\sqrt{x^2+y^2}}\le(x^2+y^2)^{3/2}, $$ since $|x|\le\sqrt{x^2+y^2}$ and $|y|\le\sqrt{x^2+y^2}$. The right-hand side $(x^2+y^2)^{3/2}$ converges to zero when $x+iy\to0$. So $f$ is differentiable at $0$ and $f'(0)=0$.


0

Suppose $z_0 \in \mathbb{C}$. We say that a function $f$ is complex differentiable at $z_0$ if $\lim_{z \to z_0} \frac{f(z) - f(z_0}{z - z_0}$ exists and is finite. Implicit in this definition is the fact that complex differentiation is not defined at a single point $z_0$ but for all points in a open set $B(z_0, r)$ about $z_0$. What I am saying is that a ...


0

Your solution is right. As you said, at $z=0$, the function $f(z)=\cos(\frac{1}{z^2})$ has an essential singularity, that is, all terms in the Laurent series have negative powers of $z$. One aproach (as I think you did), is expanding the series for $\cos(\frac{1}{z^2})$ and multiplying it by $z^7$. Then you can calculate the residue at $z=0$, which is the ...


1

$e^{i\theta}(z-u)$ is not the Distance. The distance is $\vert e^{i\theta}(z-u) \vert$. Now it should be quite simple to continue.


1

Note that if $u,v \in \mathbb{C}$ then $|u-v|=|e^{i\theta}u-e^{i\theta}v|=|e^{i\theta}||u-v|$ since $|e^{i\theta}|=1$


2

Well, I think its just a calculation mistake which you have done. Just check it out. It comes out to be $2+z$ which is a translation.


1

For real $\theta$, the complex conjugate of $e^{i\theta}$ is $e^{-i\theta}$ and the complex conjugate of $e^{-i\theta}$ is $e^{i\theta}$. If a is real, then $\overline{a}= a$.


1

Hints: Conjugation distributes over sums and products, so $\overline{a+b}=\overline{a}+\overline{b}$ and $\overline{ab}=\overline{a}\cdot\overline{b}$ If $\theta$ is real, then $$\overline{e^{i\theta}}=\overline{\cos\theta + i\sin\theta}=\cos\theta -i\sin\theta =e^{-i\theta}$$


0

You made a mistake at the beginning. One has $$\tau^{-1}(z) = e^{i\theta}\overline{z-a}.$$


0

There are some applications in high end fluid dynamics and thermo. (Not sure if you consider that physics, but at least it's not E&M. So MechE and its offshoots: nuclear, aero, etc. You might consider other areas instead (time is limited) that are more applied or that you enjoy more or are easier for you: advanced linear algebra, PDEs, abstract ...


0

Cheating a little, for $\dfrac\infty\infty$ use $$\frac{2^{S_k}}{2^{T_k}}=2^{S_k-T_k},$$ where $S$ and $T$ are the sequences as defined above. For $0\cdot\infty$, use $$2^{-T_k}\cdot2^{S_k}=2^{S_k-T_k},$$ and for $\dfrac 00$, use $$\frac{2^{-T_k}}{2^{-S_k}}=2^{S_k-T_k}.$$


0

The best book (in my opinion) to learn properly the $\delta-\varepsilon$ definitions of continuity (and convergence of sequences etc) is Michael Spivak, Calculus.


3

Here is an answer close to the proof you have for $\infty-\infty$. Take $(0,\frac12,0,\frac14,0,\frac16,0,\ldots)$ as a representative of $0$ and let $(1,2,3,4,5,6,7,\ldots)$ represent $\infty$. If you multiply one by the other (term-by-term) you get $(0,1,0,1,\ldots)$, and this shows that $0\cdot \infty$ is nonsensical. For $\frac00$, keep the first ...


2

{$1/n$} and {$1/n^2$} tend to 0. {$n$} and {$n^2$} tend to $\infty$. Let's compare: {$(1/n)/(1/n^2) = n$}$\rightarrow \infty$ so $0/0 $~$ \infty$. {$(1/n^2)/(1/n) = 1/n$}$\rightarrow 0$ so $0/0 $~$ 0$. {$(n^2)*(1/n) = n$} so $\infty * 0 $~$\infty$ {$(n)*(1/n^2) = 1/n$} so $\infty * 0 $~$0$. {$n^2/n = n$} so $\infty / \infty $~$\infty$ {$n/n^2 = 1/n$} ...


0

Of course you could arbitrarily define a multiplication on $\overline{\mathbb{C}}$ such that (for example) $0.\infty = 35$. But a "good" multiplication on $\overline{\mathbb{C}}$ is expected to be continuous and to extend the traditionnal complex multiplication. One can see that this is impossible to define because, if $z \to \infty$, then $z \to \infty$ ...


0

The problem with $\dfrac 5 0$ is that the equation $0\cdot x=5$ has no solution for $x$. The problem with $\dfrac 0 0$ is that the equation $0\cdot x=0$ has many solutions for $x$. Things like $\dfrac 6 2$ are not problematic because the equation $2\cdot x = 6$ has just one solution for $x$. Convergence: As far as convergence goes, not that $\dfrac{6x} ...


1

Your answer is true but you could go faster with polar form. If $z = re^{i\theta}$ with $r<1$ and $\theta \in ]0,\pi[$ then $\frac{1}{z} = \frac{1}{r} e^{-i\theta}$ where $\frac{1}{r} > 1$ and $-\theta \in ]-\pi,0[.$


2

Your function is analytic on its domain. It’s well-known that an analytic function that takes only real values must be constant. Why? Nonconstant analytic functions are open mappings, i.e. must map open subsets of the domain to open subsets of the target space.


2

It represents $\frac{\partial \phi}{\partial \bar z}$, the Wirtinger derivative defined by $$ \frac{\partial \phi}{\partial \bar z} := \frac 12 \left(\frac{\partial \phi}{\partial x} + i\frac{\partial \phi}{\partial y}\right) $$


0

No! It would be if $\frac{\partial Q}{\partial x}= \frac{\partial P}{\partial y}$, not "-". You could also check that directly. For example, if you go from z= 0 to z= 1+ i ((0, 0) to (1, 1)) along the straight line y= x, the integral is $\int_0^1 (4x^2- 4x^2+ 2x)dx+ (8x^2- 2x)dx= \int_0^1 8x^2dx= \frac{8}{3}$. But if you integrate along the line from z= ...


1

The condition is equivalent to $\operatorname{Re} f\le K|z|^m$ for some $m$ and $K$. Under this condition we can conclude that $f(z)$ is a polynomial of order less than or equal to $m$. Let $f(z)=u+iv=\sum_{k=0}^\infty a_kz^k$ and $A(r)=\max _{|z|=r} u(z)$. It is well-known that for $k\ge 1$ $$ a_kr^k=\frac{1}{\pi}\int_0^{2\pi} u(re^{i\theta ...


0

For any $a\neq 0$, we can define a local branch of $\log$ in a small ball $U$ around $a$ and define $$g(z)=f\left(e^{\frac{\log z}{n}}\right)$$ for $z\in U$. Furthermore, the function $g$ is actually independent of the branch of $\log z$ chosen, since any different branch just multiplies $e^{\frac{\log z}{n}}$ by some power of $\xi$. Thus these local ...


6

Using power series is cheating because it makes it too easy. Say $$f(z)=\sum_{k=1}^\infty a_k z^k.$$ Now $f(\xi z)=f(z)$, with uniqueness of the power series coefficients, shows that $$\xi^ka_k=a_k$$for all $k$. If $k$ is not a multiple of $n$ then $\xi^k\ne1$, so $a_k=0$. Hence $$f(z)=\sum_{j=0}^\infty a_{jn}z^{jn}=g(z^n)$$if $$g(z)=\sum_{j=0}^\infty ...


1

equality holds if and only if either $z$ or $w$ is a real multiple of another You just proved that $c$ needs to be $\ge 0$ when you choose the $-$ sign in $|z \pm w|$. Similarly $c \le 0$ will work for the $+$ alternative. So for an arbitrary $c \in \mathbb R$ at least one of the inequalities will become an equality. Granted, the wording of the problem ...


1

Suppose $\alpha\in\mathbb{C}$; we want to show $\alpha$ is not a limit point of $A$. Since $A$ is finite, you can consider $$ \varepsilon=\min\{|\alpha-x|:x\in A, x\ne\alpha\} $$ Then $D_\varepsilon(\alpha)\cap A\subseteq\{\alpha\}$. Indeed, if $x\in D_\varepsilon(\alpha)\cap A$, then $|\alpha-x|<\varepsilon$, so the only possibility is $x=\alpha$. ...


2

Hint: Divy the plane up into three pieces in a clever way. This will divy $\{1,2,\dot,n\}$ into sets $E_1,E_2,E_3$ based on where the $z_i$ are located. Pick the $E_i$ to maximize $\sum_{j\in E_i} |z_i|$ and call it $E$. Then you'll see that: $$\sum_{i\in E}|z_i| \geq \frac{1}{3}\sum_{i=1}^{n} |z_i|$$ Now if you picked your three pieces cleverly, you will ...


0

Yes, you're correct. The function must by the identity theorem coincide on the domain of $f$. And there's no entire function such that $g(z)=1/z$ for $z\ne 0$. To see this you use the fact that $g$ has to be continuous ond therefore $g(0)=\infty$, but it has to be derivable there which would not work.


2

Your argument is incomplete, but can be improved by using the identity theorem to deduce that for any such entire function $g$, it must satisfy $g(z) = \frac{1}{z}$ for $z\neq 0$. However, it still needs to be shown that no such $g$ exists (maybe you've done this already, in which case, you have the result). One way to prove this is to compute an integral ...


2

Let $C$ be the circle centered at $0$ and of radius $2$ (which is included in the domain of $f$). Now, compute: $$\oint_C f(z)\,\mathrm{d}z=2i\pi\neq0.$$ Hence, by Goursat's Theorem (aka Cauchy's Integral Theorem), there are no entire functions $g$ such that $f$ and $g$ coincide on $C$.



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