# Tag Info

2

This exercise is solved using Cauchy's Integral Formula: $$f^{(m)}(0)=\frac{m!}{2\pi i}\int_{|z-z_0|}\frac{f(z)\,dz}{z^{m+1}}$$ and hence, for $m\ge n$: \begin{align} \lvert f^{(m)}(0)\rvert&=\frac{m!}{2\pi}\left|\int_{|z-z_0|=R}\frac{f(z)\,dz}{z^{m+1}}\right| \le \frac{m!}{2\pi}\cdot \frac{2\pi R}{R^{m+1}}\max_{|z|=R}\lvert f(z)\rvert ...

0

The equation to solve is thus $f^5(z)-z=z^{32}-z=0$. Which gives $z=0$ and the unit roots of degree $31$. So you have to find all $k$ such that $$\frac{2\pi}{31}k\in[-0.74 , -0.44].$$ This can be transformed into an interval for $k$. For this point to be unstable you need $1<|f'(z)|=32\cdot |z^{31}|$, which obviously is satisfied for all unit roots.

0

Step 1.Similarly as in Proposition 1,$Z_f$has no limit points on $D'\times \partial D_n$.Let $\zeta'\in D'$,let $\alpha_1(\zeta'),\ldots,$$\alpha_m(\zeta')be the zeros of f(\zeta',z_n) in D_n,and let U_{nj}:=\{|z_n-\alpha_j(\zeta')|\le r_{nj}\} are pairwise nonintersecting disks in D_n.Since Z_f is closed in D,there exists a polydisk ... 1 If f(z) were differentiable, then f(z)-3z^2=\bar{z} would be differentiable since 3z^2 is. However, \bar{z} is not complex differentiable. Check this with the Cauchy-Riemann equations. -1 This function is differentiable everywhere except at z=0. As with real differentiation, complex differentiation is still a linear operation, meaning that d/dz( 3z^2 + \bar{z}) = d/dz (3z^2) + d/dz (\bar{z}). The first term is differentiable everywhere. The second term is not differentiable at zero because \bar{z}'(0) = \lim_{z->0} ... 2 Absolute value of polynomial tends to infinity for \left|z\right|\to\infty. That is, for each M>0, there exists R>0 such that for \left|z\right|>R we have \left|p(z)\right|>M. Take sufficiently large closed disk, so that \left|p(z)\right|>1 for z outside the disk. The disk is compact, so it's image by \left|p(z)\right| is ... 0 Set$$ f(z)=\exp(\pi z/2), $$Then f'(z)\ne 0, f is one-to-one on$$ S=\{z\in\mathbb C : 0<\mathrm{Im}\, z<1\}, $$and$$ f[S]=Q=\{z\in\mathbb C: \mathrm{Re}\,z>0,\,\mathrm{Im}\,z>0\} $$The harmonic function on S is v=\frac{\pi}{2}\mathrm{Im}\, z and u=v\circ f^{-1} which is constant on the boundaries of Q. In fact ... 5 It depends on what is meant by "polynomial". If only \sum c_n z^n, then every function that is uniformly approximable by polynomials must be holomorphic on the interior of J. Although that condition is trivially satisfied if J has empty interior, that doesn't mean that for such J every continuous function is the uniform limit of polynomials. For ... 1 Not exactly. Assuming you mean that the curve is given by the equation \phi(x,y)=0, where \phi is some real valued function in the plane... then the gradient, \nabla\phi=(\phi_x, \phi_y), is normal to the curve. This can be rewritten in complex notation as \phi_x+i\phi_y, which is, by definion, 2\partial\phi/\partial\bar{z}. Now ... 2 Look at the terms of the series. Ignoring the (-1)^n for the moment, we have$$\frac{n^{2x-1}-1}{n^x}n^{iy} = n^{x-1}n^{iy} - n^{-x}n^{iy} = n^{z-1} - n^{-\overline{z}}.$$The first term is holomorphic, and hence harmonic. The second term is antiholomorphic, and hence harmonic. Thus the difference of the two terms is harmonic. So every term in the ... 0 This question illustrates an issue with the \cdots notation; it's not necessarily clear how to interpret it for small n, when there are more terms explicitly written out in the general formula than you actually want. To be clearer, you might write the same formula using the more compact summation notation:$$(\cos x+i\cdot \sin x)^n = \sum_{j=0}^n ... 0 $$(\cos x+i\sin x)^n=\sum_{k=0}^{n}\binom{n}{k}(\cos x)^{n-k}(i\sin x)^k=$$ $$=\binom{n}{0}(\cos x)^{n}(i\sin x)^0+\binom{n}{1}(\cos x)^{n-1}(i\sin x)^1+...+\binom{n}{n}(\cos x)^{0}(i\sin x)^n=$$ $$=(\cos x)^{n}+in(\cos x)^{n-1}\sin x+...+i^n(\sin x)^n$$ 0 Think about this. In the definition of derivative of a function$f$at a point$x_0$: $$\lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h},$$ what doesn't make sense if$f:\mathbb R^2\rightarrow\mathbb R^2$but does if$f:\mathbb C\rightarrow\mathbb C$? 1 Topologicaly they are the same, but algebraicaly not. Elements in$R^2$and$C$have different rules. Example in$R^2$we have$(x_1,y_1)(x_2,y_2)=x_1x_2+y_1y_2$, but in$C$we have$(x_1+y_1i)(x_2+y_2i)=x_1x_2-y_1y_2+(x_1y_2+x_2y_1)i$or$(x_1,y_1)(x_2,y_2,)=(x_1x_2-y_1y_2,x_1y_2+x_2y_1)$. 2 Hint: You can write the series as $$\sum_{k=1}^\infty\frac1{z+2k}-\frac1{z+2k-1}=-\sum_{k=1}^\infty\frac1{(z+2k)(z+2k-1)}$$ which does converge absolutely. 0 Yes. This is easy to see once you realize that projective transformations in$\mathbb{P}^n$are given by linear transformations in$\mathbb{C}^{n+1}$by thinking of projective space as the subspace of points where the$n+1$-st coordinate equals$1(plus the stuff at infinity). It's not clear to me exactly what definition you are using but from what I can ... 1 \begin{align} \operatorname{Res}\limits_{z=1/2} \frac{z^4+1}{z^2(z-2)(1-2z)} &= \lim_{z\to 1/2} (z-\frac12) \cdot \frac{z^4+1}{z^2(z-2)(1-2z)} \\ &= \lim_{z\to 1/2} \frac{z^4+1}{z^2(z-2)\cdot(-2)} = \frac{(1/2)^4+1}{(1/2)^2\cdot(-3/2)\cdot(-2)} = \frac{17}{12} \end{align} 3 When0<x<1$, we have the identity$K(x^{-1})=x(K(x)-iK(\sqrt{1-x^2}))$. Therefore $$\int^{\infty}_{1}K^2(x)\frac{dx}{x}=\int^{1}_{0}K^2(x^{-1})\frac{dx}{x}\\ =\int^{1}_{0}\left(x(K(x)-iK(\sqrt{1-x^2}))\right)^2\frac{dx}{x}\\ =\int^{1}_{0}x\left(K(x)-iK(\sqrt{1-x^2})\right)^2dx\\ =\int^1_0(xK^2(x)-xK^2(\sqrt{1-x^2})-2ixK(x)K(\sqrt{1-x^2}))dx.$$ We ... 0 Put$z= e^{ix}$to evaluate$\int^{2\pi}_0\frac{\cos(2x)}{5-4\cos(x)}dx$we evaluate$\int^{2\pi}_0\frac{(e^{ix})^2}{5-4\cos(x)}dx$,after you evalue this, its real part will be the solution of required integral now,this becomes$\int^{2\pi}_0\frac{\cos(2x)}{5-4\cos(x)}dx$=Re{$\int^{2\pi}_0\frac{(e^{ix})^2}{5-4\cos(x)}dx$} and ... 0 I think you need to assume that$C$is a simple closed curve. Otherwise, take a long skinny ellipse (a "sausage") and bend it until the two ends just touch, like your thumb touching your forefinger when you use them to make a circle. There's clearly a curve that bounds the sausage (even after the ends touch at a single point), and this curve is clearly ... 0 I have to rewrite the whole what I said: I think you need some conditions on$C$and$E$to avoid pathology and to say something useful. The conditions that come to my mind include: 1)$E$is path-connected, (since$E$is a domain, it is open and connected but I don't think they are strong enough to prevent pathological phenomena to take place.) 2) the ... 1 In general, $$\int_\gamma f(z)\ dz = \int_0^1 f(\gamma(t)) \gamma'\ dt$$ where$\gamma$is a parameterized curve between$A$and$B$such that$\gamma(0) = A$and$\gamma(1) = B$. Choose any such$\gamma$that makes your life easy. Perhaps a straight line between$A$and$B$will work: $$\gamma(t) = (1-t)A+tB.$$ The next step: $$\gamma'(t) = B-A.\\ ... 2 Note that \operatorname{re} f'(z) >0 for all z \in B(0,1). Suppose f(z_1) = f(z_2) with z_1 \neq z_2. Let \gamma:[0,1] \to B(0,1) be given by \gamma(t) = z_1+t(z_2-z_1), then f(z_2)-f(z_1) = 0 = \int_0^1 (f \circ \gamma)'(t) dt = \int_0^1 f'(\gamma(t)) (z_2-z_1) dt = \left( \int_0^1 f'(\gamma(t))dt \right) (z_2-z_1), from which we get ... 2$$ z^2-\sqrt{2}\,z+2=(z-a)(z-b)\text{ where }a=\frac{\sqrt2+\sqrt6\,i}{2},\ b=\frac{\sqrt2-\sqrt6\,i}{2}. $$Use partial fraction decomposition to write$$ \frac{1}{z^2-\sqrt{2}\,z+2}=\frac{A}{z-a}+\frac{B}{z-b}. $$Now expand in power series 1/(z-a) and 1/(z-b). 0 I think your answer's correct, but why not going directly by the very definition and by the main logarithmic branch?:$$\text{Log}\,(1-\sqrt3\,i):=\log|1-\sqrt3\,i|+i\arg(1-\sqrt3\,i)==\log2+i\arctan\frac{-\sqrt3}1=\log 2-\frac\pi3i\implies\text{Log}^i\,(1-\sqrt3\,i)=e^{i\,\text{Log}\,(1-\sqrt3\,i)}=e^{\frac\pi3+i\log2}$$0 If you really mean f(z)=\text{Log}(z+2), then you can use the CG-theorem. Of course, the function \text{Log}(z) has a branch cut along the negative real axis. But if you look at the function \text{Log}(z+2) the branch cut starts at z=-2, so \text{Log}(z+2) is analytic in the disc with radius one and you can use the CG-therom to show that the ... 1 \lim\frac{|a_{n+1}z^{n+1}|}{|a_nz^n|}=\lim\frac{|a_{n+1}||z^{n+1}|}{|a_n||z^n|}=\lim\frac{|a_{n+1}|}{|a_n|}|\frac{z^{n+1}}{z^n}|=\lim\frac{|a_{n+1}|}{|a_n|}|z|=|z|\lim\frac{|a_{n+1}|}{|a_n|} 1 These inequalities are addressed in \S 11.5.3 of these notes. (More precisely the middle inequality is clear, and the two outer inequalities are very similar, so the notes carefully prove one of them.) As Daniel Fischer has pointed out, the implication you ask about at the end of your question is false in general. What is true is \lim_{n \rightarrow ... 1 Not quite. In step 2., you have$$\limsup_{n\to\infty} \lvert a_nb_n\rvert^{1/n} = \limsup_{n\to\infty} \bigl(\lvert a_n\rvert\,\lvert b_n\rvert\bigr)^{1/n} \leqslant \limsup_{n\to\infty} \lvert a_n\rvert^{1/n}\cdot \limsup_{n\to\infty} \lvert b_n\rvert^{1/n}$$instead of what you wrote, and correspondingly in step 3. you also need to swap the = and ... 1 If \liminf \frac{a_{k+1}}{a_k} = 0, the first inequality is clear. Thus let us suppose it is strictly positive. Let 0 < c < \liminf \frac{a_{k+1}}{a_k}. Then there is a K_c such that for all k \geqslant K_c we have \frac{a_{k+1}}{a_k} > c, and hence$$a_{K_c+n} = a_{K_c}\prod_{j=0}^{n-1} \frac{a_{K_c+j+1}}{a_{K_c+j}} > a_{K_c}\cdot ... 0 in power series $$\cos z=\sum\limits_{n=0}^\infty \frac{z^{2n}(-1)^{n}}{(2n)!}$$ and cosine function is continuous every where 1 An alternative, and imo much simpler, way. Put $$a_n:=\frac{z^{2n}(-1)^n}{(2n)!}\implies\;$$ $$\frac{a_{n+1}}{a_n}=\left|\frac{z^{2n+2}(-1)^{n+1}}{(2n+2)!}\cdot\frac{(2n)!}{z^{2n}(-1)^n}\right|=|z|^2\frac1{(2n+1)(2n+2)}\xrightarrow[n\to\infty]{}0$$ and thus the series' radius of convergence is infinite, which means the series defines an analytic function ... 1 We don't need Rouché's theorem to see the continuity of$a(z')$. The continuity - and even the holomorphy - of$a$follows by the residue theorem, which yields the representation $$a(z') = \frac{1}{2\pi i}\int_{\partial U_n} \zeta\frac{\frac{\partial f}{\partial z_n}(z',\zeta)}{f(z',\zeta)}\,d\zeta,\tag{1}$$ and standard results of integration theory (the ... 1 What is going on is that the complex numbers are being considered as vectors in a two-dimensional plane. You can see that geometrically an X-Y plane represents a complex number of the form x + iy as the point (x, y), or equivalently by the vector connecting the origin to the point (x, y). The angle between two vectors in a two-dimensional plane is given by ... 1 The last formula is wrong; there should be a complex conjugate on either$z_1$or$z_2$. The formula should read$v_1\cdot v_2=\Re(z_1 \overline{z_2})=\Re(\overline{z_1} z_2)$. Now, you can verify the formula by expanding out both sides in terms of$a$and$b$. 0$z_1=a_1+ib_1$and$z_2=a_1+ib_1z_1z_2=(a_1+ib_1)(a_2+ib_2)=a_1a_2+i(a_1b_2+a_2b_1)-b_1b_2Re(z_1z_2)=a_1a_2-b_1b_2$0 Since there is a sequence$z_k$tending to infinity for which$\sin z_k = 0$for every$k$, the corresponding sequence$1/z_k$shows that$\sin(1/z)$does not tend to infinity at the origin - hence, no pole. Alternative way: Show that a function has a pole at infinity if and only if it's a non-constant polynomial. This is a good exercise in Liouville's ... 1 The singularity at$\infty$is not a pole: it is essential. This is a fact that is easy to prove: consider the limit $$L(\theta) = \lim_{r \to 0^+} |\sin(re^{i\theta})^{-1}|.$$ For$\theta = \pi k$for some integer$k$, this limit is bounded in the interval$L(\theta) \in [0,1]$. If$\theta$is not an integer multiple of$k$, this limit is$\infty$. 0 No. Take$z_n=-1$, so that$\prod_{n=1}^\infty |z_n|$trivially converges to$1$. But$z_n=1+(-2)$, and the definition of$\prod_{n=1}^\infty (1+(-2))$converging absolutely is that$\prod_{n=1}^\infty (1+|{-}2|) = \prod_{n=1}^\infty 3$converges, which it does not. 0$\displaystyle\sin z=-\frac54\displaystyle\cos z=\pm\sqrt{1-\frac{25}{16}}=\pm\frac{3i}4$Using Euler formula, $$e^{iz}=\cos z+i\sin z=\pm\frac{3i}4+i\left(-\frac54\right)=-\frac{(5\mp3)i}4=-2i\text{ or }-\frac i2$$ 2 To start, recall that$\sin{z} = \frac{1}{2i}(e^{iz}-e^{-iz})$. So we are looking for$z$such that $$e^{iz}-e^{-iz} = -\frac{5i}{2}.$$ Multiplying both sides by$e^{iz}$and bringing all terms to the left yields $$(e^{iz})^2+\frac{5i}{2}e^{iz}-1=0.$$ Setting$w = e^{iz}$, we have the quadratic ... 0 $$\sin a=\frac12=\sin\frac\pi6\implies a=n\pi+(-1)^n\frac\pi6$$ where$n$is any integer As "a belong to the first positive quadrant,"$\displaystyle a=\frac\pi6$(setting$n=0$) 1 Part 1 I don't think your answer to this is correct. You need to justify why we can be sure that$g$has a power series that converges if$|z|<1$, and you haven't done that. You have correctly shown that$g$is a product of two power series: $$g(z)=\sum_{k=0}^{\infty}a_kz^k *\sum_{n=0}^{\infty}z^n$$ We also know that$f(z)=\sum_{k=0}^{\infty}a_kz^k$has ... 0 Thanks to the Cauchy-Riemann equations we have: $$u_x=v_y=2uu_y,\ u_y=-v_x=-2uu_x.$$ It follows that $$u_x=2uu_y=-4u^2u_x,\ u_y=-4u^2u_y,$$ i.e. $$(1+4u^2)u_x=0=(1+4u^2)u_y.$$ Hence $$u_x=u_y=v_x=v_y=0,$$ and since$D$is connected, it follows that$u$and$v$are constant. Thus$f$is constant. 1 Let's apply Cauchy-Riemann: $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$ says that $$\frac{\partial u}{\partial x} = \frac{\partial u^2}{\partial y} = 2u\frac{\partial u}{\partial y}.\tag{1}$$ Likewise $$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$ says that $$\frac{\partial u}{\partial y} = -\frac{\partial ... 1 Hint:$$u_x = v_y = (u^2)_y = 2uu_y$$Now use the other Cauchy-Riemann equation to represent u_y in terms of u and u_x, and rearrange. Spoiler: 1 Part of the problem is that z\mapsto a^c is "multiple-valued". The simplest example is when c=1/2. You learned at your mother's knee that each positive real number z has two square roots, called \pm\sqrt{z}, and each negative real number z has two square roots, called \pm i\sqrt{-z} (where of course -z is a positive number, and so is ... 0 e^{\cos\theta+i\sin\theta} can be explained as follows. i) e^{i\theta} is a rotation about \theta radians in the complex plane. e^{\cos\theta+i\sin\theta}=\exp({e^{i\theta}}) is the a rotation in \mathbb{C}^1 represented in the space X, where \exp:\mathbb{C}^1\to X. ii) The range of \cos\theta, and \sin\theta is [-1,1]. So, as you've ... 1 Use d'Alembert rule: Let z\in\mathbb C.$$ \frac{|a_{n+1} z^{n+1}|}{|a_nz^n|} \to \frac{|z|}{R} $$so when |z|<R there is convergence, and when |z|>R there is divergence. Hence, R is the radius. 1 Notice that you are allowed to use the relation \exp(z+w) = \exp(z)\exp(w). Use also (iii) and just notice that$$ 1 = \exp(0) = \exp(z+(-z)) = \exp(z)\exp(-z).$\$

Top 50 recent answers are included