# Tag Info

1

Given two points $a, b \in G \backslash f^{-1}(0)$, start with the straight line segment joining them. This may contain points of $f^{-1}(0)$, but only finitely many. Make a sufficiently small semicircular detour around each of them.

2

You can just replace $G$ by $G\setminus g^{-1}(\{0\})$, and now you've reduced to the case where $g$ has no zeroes in $G$.

2

Yes, it holds in general. If you let $G_\epsilon$ be the open subset of $G$ which consists of the points $z$ with $|1/z|<\epsilon$ and $\textrm{dist}(z,\partial G) > \epsilon$, then for $\epsilon > 0$ sufficiently small, $\gamma \subset G_\epsilon$, and $f$ has finitely many zeros in $G_\epsilon$ because it is compactly contained in $G$. In order to ...

2

The trace of $\gamma$ is not necessarily contained in the closure of $H$. A simple example is $$\gamma(t) = \bigl\lvert \tfrac{1}{2} - t\bigr\rvert.$$ This is a rectifiable (it's piecewise continuously differentiable even) closed curve, and $H = \varnothing$ for this curve. More generally, if $\gamma$ enters into a region where the winding number is $0$ ...

0

We may assume without loss of generality that $\alpha =f(0)$ is real positive. Fix $r<1$. We prove that$$\left|f(re^{i\theta })-\frac{\alpha (1-r^2)}{1-\alpha ^2r^2}\right| \le \frac{r(1-\alpha ^2)}{1-\alpha ^2r^2}\tag{1}$$for $0\le \theta <2\pi$. Let $$F(z)=\frac{f(z)-\alpha }{1-\alpha f(z)}.$$ Since $F(0)=0$ and $|F(z)|< 1$, $|F(z)|\le |z|$ by ...

2

The analytic continuation theorem states that two holomorphic functions on a domain which agree on an open subset of that domain are equal. (The statement can be made stronger but we don't need more.) Therefore any function $f$ which is holomorphic on your disk and satisfies $f(z)^3 = e^z$ is actually equal to $e^{z/3}$ on all of $\mathbb C$ (see Edit : or ...

1

If $f(z) = z^z$, then $\log f(z) = z \log z$, so differentiating implicitly with respect to $z$ yields: $$\frac{f'(z)}{f(z)} = 1 + \log z.$$ Multiplying through by $f(z)$ yields $$f'(z) = z^z(1 + \log z).$$

0

Hint: Use logarithmic differentiation. Can you take it from here?

0

$$log(z^\frac{p}{q}) = \frac{p}{q}log(z) = \frac{p}{q}(Log(z) + i2\pi k) + i2\pi n$$ The justification I recall is that the exponential of the LHS equals that of the RHS. The derivation went along the lines of: Replace $z$ with $ze^{i2\pi k}$. Let $y=(ze^{i2\pi k})^\frac{p}{q}$. $$log(y) = Log(y) + i2\pi n = Log((ze^{i2\pi k})^\frac{p}{q}) + i2\pi n = ... 1 \bullet Since you want to check that f'(z) does not exist at z=0, seems to me that the problem is to conclude that the Cauchy-Riemann equations hold, but only at z=0, as @Dr.MV pointed out in the comments. To see that lets put f(x+iy)=u(x,y)+iv(x,y), hence$$ \frac{\partial u}{\partial x}(0,0)= \lim_{x \to 0} \frac{u(x,0)-u(0,0)}{x} = \lim_{x \to ...

0

Yes: we have $$\int_D f(\lvert z \rvert ) \, dA = \int_0^1 \int_0^{2\pi} f(r) r \, d\theta \, dr$$ by Fubini's theorem. Then $$0 \leqslant \int_D f(\lvert z \rvert ) \, dA = 2\pi \int_0^1 r f(r) \, dr \leqslant 2\pi \int_0^1 f < \infty$$

0

Integrate in polar coordinates and you get $$\int_{\Bbb D}f(|z|)\,dA=2\pi\int_0^1r\,f(r)\,dr\le 2\pi\int_0^1 f.$$

0

This is analogous to: $$\cos 0 = \cos{2\pi}\implies \pi = 0$$

3

The second equality is not true in $\mathbb{C}$, but only in $\mathbb{R}$. See this.

-1

That's right, $i^i$ has many values and all of them are real. There are many counterintuitive things about complex numbers. A function of square root, for example, always has two values, and we have no reason to say that one of them is "the true one" and the other is not. Cube root has 3 values. Logarithm has infinitely many values (unless you define a ...

1

The mapping $z \mapsto z^\alpha$ is defined by $f_{\alpha}(z) = e^{\alpha\log(z)}$, note that the complex logarithm is defined with $\log(z) = \log |z| + i\arg(z)$ That being said, we can calculate: $i^i = e^{i\log(i)} = e^{i ( \log|i| + i\arg (i)) } = e^{-1\cdot(\frac{\pi}{2} + 2\pi k)}$. There are infinitely many results. This is due to the ...

0

First draw the original locus, which is a part circle on the chord connecting the points $z=3$ and $z=i$ and extending into the first quadrant, so that every point on the arc subtends the angle $\frac{\pi}{6}$ with the chord. The reflection in the real axis will be an arc in the fourth quadrant based on the chord connecting $z=3$ and $z=-i$, so, when ...

1

$log(x) = Log(x) + i2\pi n = Log(|x|) + iArg(x) + i2\pi n$ where $Log$ and $Arg$ are principal values. $log(x) + log(x) = 2Log(|x|) + i2Arg(x) + i2\pi n + i2\pi m$ $n$ and $m$ are independent and can be merged into one variable $k$. $alog(x) = aLog(|x|) + iaArg(x) + i2\pi k$ $$alog(x) = aLog(x) + i2\pi k\ \ \ (1)$$ $e^{alog(x)} = e^{aLog(x) + i2\pi k} = ... 0 HINT 1: $$i=e^{i(\pi/2+2\ell \pi)}$$ HINT 2: $$\sqrt{i}=\pm e^{i\pi/4}=\pm \left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)$$ 0 This follows from the proof of Dirichlet's test, once we observe that for$|z| \le 1, |z-1|\le r,$the following uniform bound holds: $$\tag 1 |1 + z + z^2 + \cdots + z^n| = |(z^{n+1}-1)/(z-1)| \le 2/|z-1| \le 2/r.$$ For the proof of Dirichlet's test, see here for example: https://en.wikipedia.org/wiki/Dirichlet%27s_test 3 For$0<r<2$we consider$\displaystyle D_r=\{z\in\mathbb{C}:\vert z\vert\le 1,\vert z-1\vert\ge r\}\qquad\qquad$Also, let$S_n(z)=\sum_{k=0}^na_kz^k$. Now, it is clear that for$m>nwe have \eqalign{(1-z)(S_m(z)-S_n(z))&=(1-z)\sum_{k=n+1}^ma_kz^k\cr &=\sum_{k=n+1}^ma_kz^k-\sum_{k=n+2}^{m+1}a_{k-1}z^k\cr ... 0 For any integer r,\left(e^{2r\pi i/n}\right)^n=e^{2r\pi i}=1$$But if f(z)=z^n-1 f'(z)=nz^{n-1} Hence f(z) can not have repeated roots Again if e^{2a\pi i/n}=e^{2b\pi i/n}\iff e^{2(a-b)\pi i/n}=1 \iff n|(a-b)\iff a\equiv b\pmod n So, the roots of z^n=1 are e^{2r\pi i/n}, r\equiv0,1,\cdots,n-2,n-1\pmod n 0 Hint$$z^n=1\iff z=e^{\frac{2 ik\pi}{n}}$$with k=0,...,n-1. 1 You can find a proof in Chapter IV, Section D, Theorem 2 of Gunning and Rossi, Analytic functions of several complex variables. They prove, more strongly, that if f_{\alpha} is any set of holomorphic functions on U, p is any point of U and m \geq 0 is any integer, then there is an open neighborhood U' of p on which the ideal (f_{\alpha}) has ... 1 First apply the formula for the sum of a geometric progression(assume \phi\ne 0 ):$$1 + e^{i \phi} + e^{2 i \phi} + \ldots + e^{i n \phi}=\frac{e^{i(n+1)\phi}-1}{e^{i\phi}-1}$$Break the numerator and denominator like this:$$\frac{e^{i(n+1)\phi}-1}{e^{i\phi}-1}=\frac{e^{i\phi(n+1)/2}}{e^{i\phi /2}}\cdot\frac{e^{i\phi(n+1)/2}-e^{-i\phi(n+1)/2}}{e^{i\phi ... 0 Hint: Geometric series: $$\sum_{k=0}^n q^k = \frac{1-q^{n+1}}{1-q}$$ forq\neq 1$1 Take $$x_n =\left( 1 +\frac{1}{n} , \frac{1}{2} +\frac{1}{n} ,...,\frac{1}{n} +\frac{1}{n} ,0,0,...\right)$$ 1 You are correct that the phase (of the Fourier transform!) is$\pm\pi/2$, because the Fourier transform is purely imaginary (due the time signal being an odd function of time). Note that for$\omega=0$the phase in not zero, but it is undefined because the Fourier transform is zero for$\omega=0$. If you have a complex number$z=re^{i\phi}$and you set ... 0 Answer is in your question itself. The map is $$z\mapsto [z,z_2,z_3,z_4]=\frac{...?...}{...?...}$$ Therefore,$z_1\rightarrow ...?...$and so on. 0 You can parametrize$\gamma (t)=1+it$as the line going from$1$to$1+i$then use$\int_{\gamma} f(z)=\int_a^b f(\gamma (t))\gamma'(t) dt$Using the above parametrization you get$\int_0^1 \frac {idt}{1+(1+it)^2}=\arctan(1+i)-\pi/4$You can get$\arctan(z)=\frac 12 i(\ln(1+iz)-\ln(1-iz))$but that sounds harder than just putting$\arctan(1+i)$into a ... 0 If$A$is a real matrix which has complex eigenvalue$\lambda$and eigenvector$v$(i.e.,$Av = \lambda v$), then simple application of complex conjugation shows that$(\overline{\lambda}, \overline{v})$is an eigenpair. 1 It is not true. For example, let A be the identity, and any two vectors will satisfy the equation, since for the identity, every vector is an eigenvector with eigenvalue 1. 0 Your interpretation of the described signal/function is wrong. This signal has 0 phase every where.$f(t)=Asin(ω_0 t + \phi)$is a general description of a signal with phase$\phi$, and when this phase equals 0, you get a signal with phase 0. 1 Notice, we have $$\cos z=\sqrt2$$ $$\frac{e^{iz}+e^{-iz}}{2}=\sqrt2$$ $$\frac{e^{2iz}+1}{2e^{iz}}=\sqrt2$$ $$e^{2iz}+1=2\sqrt2e^{iz}$$ $$(e^{iz})^2-2\sqrt2e^{iz}+1=0$$ Above is the quadratic equation in terms of$e^{iz}$hence using quadratic formula $$e^{iz}=\frac{2\sqrt2\pm\sqrt{(2\sqrt2)^2-4(1)(1)}}{2(1)}$$$$e^{iz}=\sqrt 2\pm 1$$ $$iz=\ln(\sqrt ... 1 You have$$e^iz + e^{-iz} = 2\sqrt{2}. $$so$$e^{2iz} + 1 = 2\sqrt{2}e^{iz}. $$Put w = e^{iz}; this becomes$$w^2 + 1 = 2\sqrt{2} w.$$This is a quadratic. 6 \DeclareMathOperator{\Log}{Log} Here is what Churchill, Brown, Verhey say about this particular case (Complex Variables and Applications, p. 66, Third Edition) and I quote: "The statement \log(z^n)=n\log(z) may or may not be true for specific values of z and n when the multiple-valued complex logarithmic function is replaced by a single branch of ... 0 Maximising the absolute value is the same as maximising the square of the absolute value. Using that \lvert w \rvert^2 =\bar{w}w , we have$$ \left\lvert \frac{e^{3z}}{1+e^z} \right\rvert^2 = \frac{e^{3(z+\bar{z})}}{(1+e^z)(1+e^{\bar{z}})} = \frac{e^{6\Re{(z)}}}{1+2\Re(e^z)+e^{2\Re(z)}}, $$using that 2\Re(w) = w+\bar{w}. Now, we have \Re(z) = R, e^z ... 0 Hint: To maximize |1/f| you want to minimize |f|. As y goes from 0 to 2\pi, e^R e^{iy} goes around a circle of radius e^R centred at 0. What is the closest point to -1 on that circle? 1 If D is simply connected this follows from the Riemann mapping theorem. If not, let's see. There exists r>0 such that B(0,r)\subset D. And there exists M so |z|\le M for all z\in D, hence |f(z)|\le M. So we have |f(z)|\le M in B(0,r), hence Cauchy's estimates show there exists c such that$$|f'(0)|\le c$$for every holomorphic f:D\to ... 0 Yes if you want only one z_n for each individual zero you will need to have an exponent too. But in the text it says "number of times the linear factor appears" so I suppose they allow z_n = z_m for n\neq m. I agree your approach would be more convenient if the function had zeros of higher order than one. 0 Hint$$\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}.$$Then if you set z=e^{i\theta}$$...=\int_C\frac{dz}{iz\left(1+\frac{z+z^{-1}}{4}\right)^2}=...$$where C=\{e^{i\theta}\mid\theta\in[0,\pi]\}. 3 Hint Consider$$f(z)=\frac{ze^{iz}}{1+z^2}$$and integrate on$$\{[-r,r]\mid r>a\}\cup\{re^{i\theta}\mid \theta\in[0,\pi]\}.$$1 The function e^{senz}/|z|^3 is not analytical. This can be proven by the identity theorem: The function e^{senz}/z^3 is holomorphic on \mathbb C \setminus 0 (quotient rule) and identical to e^{senx}/x^3 for x \in (1, \infty). If the function e^{senz}/|z|^3 was holomorphic, we would have e^{senz}/|z|^3 = e^{senz}/z^3 for all z \in \mathbb C by ... 0 Hint: expanding out the polar form of \lvert z_n-z \rvert^2 gives$$ r_n^2 +r^2-2rr_n \cos{(\theta_n-\theta)}, $$if you group the terms correctly. Now find necessary conditions for this to tend to zero. 0 Just continue your approach: \cos(z) = \cos(x)\cosh(y)+i\sin(x)\sinh(y) Then just plug in z+\pi instead: \cos(z+\pi) = \cos(x+\pi)\cosh(y)+i\sin(x+\pi)\sinh(y) And observe that both \cos(x+\pi)=-\cos(x) and \sin(x+\pi)=-\sin(x) and you get: \cos(z+\pi) = -\cos(x)\cosh(y)-i\sin(x)\sinh(y) = -\cos(z) 1 Taking the parametrization \gamma(t) = it, 0 \le t \le \log 2 will be slightly less typing, so I'm choosing that one. Then \sin(it) = \frac{\exp(-t)-\exp(t)}{2i} = i \sinh t. In particular, |\sin(it)| = |\sinh t|, and you can check that \sinh is increasing on t > 0. Hence, for 0 \le t \le \log 2:$$ |\sin(it)| \le |\sin(i\log 2)| = ... 0 The sleek, more complex-analytic way to do it: Let$f(z) = \cos z + \cos(z+\pi)$. Then$f$is entire, and$f(x) = 0$for$x \in \mathbb{R}$. Hence, by the identity theorem,$f(z) = 0$for all$z \in \mathbb{C}$. 0 sum of continuous functions is continuous. product of continuous functions is continuous. x is continuous. 0 you know that if$f$and$g$be continuous then$f+g$and$f.g$are continuous. By using this, it is enough to show that$h(x)=x$is continuous. which is clear by definition of continuity. 2 Hint: Recall that$e^{\pi i} = e^{-\pi i} = -1$and that$e^{u+v} = e^{u}e^{v}\$ and go with your first way.

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