Let $\gamma$ be a continuously differentiable closed curve in the complex plane with parameter interval $[a,b]$, and assume that $\gamma (t) \ne 0$ for all $ t \in [a,b]$. Define the index of $\gamma$ to be $$\text{Ind} (\gamma) = \frac{1}{2\pi i}\int_{a}^{b} \frac{\gamma'(t)}{\gamma (t)} dt$$ Prove that this is always an integer. In the hint it's mentioned there is a function $\phi$ on $[a,b]$ with $\phi ' =\frac{\gamma'}{\gamma}$ and $\phi(a)=0$. I can understand that the primitive of $\frac{\gamma'}{\gamma}$ exists since it is continuous and integrable but cannot understand why $\phi(a)=0$ which plays a vital role in the proof.
123196
Let $f$ be holomorphic in an open $\Omega \subset \mathbb{C}$ and $\gamma$ a closed curve in $\text{int}(\Omega)$, along which $f$ is never zero.   Are these hypotheses enough to claim $\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}dz$ is an integer? If not, what are the necessary and sufficient conditions for that?  I can prove that's an integer for some particular cases, for example when $\Omega$ is convex and  $\gamma$ is a circle. But I've seen people claming this in many other contexts, like when $\Omega$ is an annulus around $0$ and $\gamma $ is an arbitrary curve inside it. This isn't obvious to me at all.  Thanks!
96766
Let $f\ge0$ be integrable on $[a,b]$. Let $f$ be continuous on $x_0\in (a,b)$ and let $f(x_0)&gt;0$.   Prove $$\int_{a}^{b}{f(x)dx}&gt;0$$  It is easy to see it, graphically, because $$f\ge 0 \Rightarrow\int_{a}^{b}{f(x)dx}\ge0$$  and if $f&gt;0$ somewhere between $a$ and $b$, then the zone of $F$ between those point is "real" (positive) and $\ne 0$.  My problem is, how do I show it formally?  I would appreciate your help.
88056
There are many theorems in complex analysis which tell us about integration $\int_{\gamma} f$ where $f$ is continuous (or even differentiable) in the interior of $\gamma$ except finitely many points.   I would like to see how should we solve $\int_{\gamma}f(z)dz$ where $\gamma$ contains a singular point of $f$.   For example, in solving $\int_{|z|=1} \frac{1}{z-1}dz$, I slightly pulled the curve $|z|=1$ near $1$ towards left, so that $\gamma$ will not contain $1$; then the singularity of $\frac{1}{z-1}$ will not be inside this deformed curve, so integration over deformed curve is zero, so taking limit, I concluded that $\int_{|z|=1} \frac{1}{z-1}dz=0$.   However, if I pull the curve near $1$ on the right side, then the singularity of $\frac{1}{z-1}$ will be inside this deformed curve, and integration is then non-zero.   I confused between these two processes; can you help me what is the correct way to proceed for finding integration along curves, in which singularity of the function is on the curve?
101590
I need to evaluate $\int_{\gamma} \frac{ |dz| }{z} $ where $\gamma $ is the unit circle.    Attempt: Let $\gamma(t) = e^{it} $ where $t \in [0,2 \pi]$. How can I get rid of the absolute values in the differential?
47843
I've been doing some more exercises in , in particular the following:  Show that $H_0(X,A) = 0$ iff $A$ meets each path-component of $X$.  "$\Leftarrow$": Let $x_i \in A \cap X_i \neq \emptyset \forall $ path-components $X_i$. Then for any $x \in X_i$ there is a path $\gamma$ from $x_i$ to $x$. They therefore differ by a boundary: $\partial \gamma = x - x_i$ and therefore $\{&#160;x \} = \{ x_i \}$ in $H_0(X,A) $. Therefore $H_0(X,A) = 0$.  "$\Rightarrow$" Now for this direction I'm not so sure and I'd be glad if you could give me a hint. Let $H_0(X,A) = 0$. By definition, the relative homology group is calculated from the following sequence of chain groups:  $$ 0 \rightarrow C_0(A) \rightarrow C_0(X) \xrightarrow{\partial_1} C_0(X) / C_0(A) \xrightarrow{\partial_0} 0$$  Then $H_0(X, A) = H_0(C_0(X) / C_0(A)) = ker \partial_0 / im \partial_0 = (C_0(X) / C_0(A) ) / im \partial_1 = 0$.  There are two cases where this equality holds:  (i) $C_0(X) / C_0(A) = 0$  (ii) $im \partial_1 = C_0(X) / C_0(A)$  In case (i), $A \cap X_i \neq \emptyset$ for all $X_i$.  In case (ii) I'm not sure how to proceed. Can you give me a hint? Many thanks for your help!  Edit Or maybe I could do the second direction like this:  $$0 = H_0(X,A) = \oplus_i H_0(X_i, A) \implies  H_0(X_i, A) = 0 \forall i$$  $$ \implies A \cap X_i \neq \emptyset \forall i$$?
47373
Let $f\in C[0,1]$ be such  that $\forall \phi\in C[0,1]$ with $\int_0^1\phi \, dx=0$,we have $\int_0^1f\phi \, dx=0$.  Then $f$ is constant.  May be someone can help me proof this problem. Thanks!
117750
Formally, if I want to measure the length of a closed curve $f(x,y) = 0$, I presumed I could write:  $$ L = \int^\infty_{-\infty}\int^\infty_{-\infty} \delta( f(x,y) )\, dx\, dy, $$   but trying this out I don't think this works. What is wrong with this formula?  Edit: Am I missing a measure like a Jacobian or something? How can you prove this?
826
Let $f: \Bbb C \to \Bbb C$ be an entire (analytic on the whole plane) function such that exists $\omega_1,\omega_2 \in \mathbb{S}^1$, linearly independent over $\Bbb R$ such that: $$f(z+\omega_1)=f(z)=f(z+\omega_2), \quad \forall\,z\in \Bbb C.$$Prove that $f$ is constant.   The intuition seems clear to me, we have the three vertices of a triangle given by $0$, $\omega_1$ and $\omega_2$. All points in the plane are one of the vertices of that triangle under a suitable parallel translation. The constant value will be $f(0)$, fine. Throwing values for $z$ there, I have found that $$f(n\omega_1) = f(\omega_1) = f(0)=f(\omega_2) = f(n\omega_2), \quad \forall\, n \in \Bbb Z.$$  I don't know how to improve the above for, say, rationals (at least). Some another ideas would be:   Checking that $f' \equiv 0$. I don't have a clue of how to do that. Write $w = a\omega_1+b\omega_2$, with $a,b \in \Bbb R$, do stuff and conclude that $f(w) = f(0)$. This approach doesn't seem good, because I only have a weak result with integers above. Finding that $f$ coincides with $f(0)$ on a set with an accumulation point. This seems also bad: the set on with $f$ coincides with $f(0)$ by which I found above is discrete.    Nothing works and this is getting annoying... And I don't see how analyticity comes in there. I'll be very thankful if someone can give me an idea.  (On a side note.. I know that this title is not informative at all. Feel free to edit if you come up with something better.)
97844
Problem : If $f(z)$ has an isolated singularity at $z=z_0$ and if $\lim_{z\to z_0} (z-z_0)^{\alpha}f(z)=M\neq0,\infty$, then prove that $\alpha$ must be an integer.  This is Exercise 9.22.1 in Silverman's "Complex Variables with Applications". Most of the exercises in this book are quite easy, but I have no idea with this question. How should I attempt?   P.S. There is no assumptions about $\alpha$, but I think it should be any complex constant.  P.P.S. When $\alpha$ is not an integer, even though I choose the principal branch of $(z-z_0)^{\alpha}$, does  $\lim_{z\to z_0} (z-z_0)^{\alpha}$ exist?
112632
My question is about the differential of the gauss map $dN_{p}$, to start the convenient expression of the symbol and formula, I have to construct some functions(maps).  First, there is a one dimension parameters line and I use the parameter as $t$, and the function $\overline{α}(t)=(u(t),v(t))$ maps $t$ to $uv$ plane, and which is a plane curve on $uv$ plane.  Second, there is a function $\mathbf{x}$ that takes $\overline{α}(t)$ to a curve on a regular surface, $α(t)=\mathbf{x}(\overline{α}(t))=(x(u(t),v(t)),y(u(t),v(t)),z(u(t),v(t)))$.  Final, there is a map $\mathbf{N}$ represents the unit normal vector on each point of the surface (x,y,z), $N(x,y,z)=(X(x,y,z),Y(x,y,z),Z(x,y,z))$.  If $p$ is a point on the surface(i.e.,$p=α(0)$),then, the linear map $dN_p$ corresponds a 3x3 matrix: \begin{bmatrix}X_x&amp;X_y&amp;X_z\\Y_x&amp;Y_y&amp;Y_z\\Z_x&amp;Z_y&amp;Z_z\end{bmatrix}  The matrix derive from the chain rule in calculus: $\frac{dN(α(t))}{dt}=(X_{x}*x'(t)+X_{y}*y'(t)+X_{z}*z'(t), Y_{x}*x'(t)+Y_{y}*y'(t)+Y_{z}*z'(t), Z_{x}*x'(t)+Z_{y}*y'(t)+Z_{z}*z'(t))=\begin{bmatrix}X_x&amp;X_y&amp;X_z\\Y_x&amp;Y_y&amp;Y_z\\Z_x&amp;Z_y&amp;Z_z\end{bmatrix}\pmatrix{x'(t)\\y'(t)\\z'(t)}=(dN)_{3x3}(α'(t))$, here, $x'(t)=\frac{dx(u(t),v(t))}{dt},y'(t)=\frac{dy(u(t),v(t))}{dt},z'(t)=\frac{dz(u(t),v(t))}{dt},$  Now ,I have the question about the partial derivative $\pmatrix{X_x\\Y_x\\Z_x}$,  the definition of it is  $$\lim_{\Delta t\to 0} \frac{(X(x+Δt,y,z),Y(x+Δt,y,z),Z(x+Δt,y,z))-(X(x,y,z),Y(x,y,z),Z(x,y,z))}{\Delta t},$$  but even $(X(x+Δt,y,z),Y(x+Δt,y,z),Z(x+Δt,y,z))$ may not defined because the domain of the map $\mathbf{N}$ is only defined on the surface, not a solid domain, so if we move a little $Δt$, it may go out of the domain(the domain is only on the surface).So, I'm not really sure how we interpret the correspondence matrix of the linear map $dN_p$, maybe there are some concepts need to be clarify, thanks for help.     Thanks Andrew and janmarqz, I think that we couldn't express the correspondence of the linear map $dN_p$ in this way(using (x,y,z) domain when we want to differentiate it at a point on the surface) because as I mention above, the partial derivative column vectors may not exist.   Instead, we can explained the $dN_p$ is in the form: $dN_{\{u,v\}} \circ \mathbf{τ_{β}}$ where $β=\{x_u, x_v\}$ is the basis of $T_{p}(S)$, and $τ_{β}$ is a linear transformation that express the vector in $T_{p}(S)$ in the basis $\{x_u, x_v\}$ as its coordinate.      $dN_{\{u,v\}}\circ τ_{β}(x_{u}*u'+x_{v}*v')=dN_{\{u,v\}}\left(\begin{array}{c} u'\\ v'\end{array}\right)=N_{u}*u'+N_{v}*v'$, so in this way, the problem is solved and $N_u$ and $Nv$ both exist because $N_u=\frac{\partial\frac{x_{u} \times x_{v}}{\|x_{u} \times x_{v}\|}}{\partial u}$, and I think that it exist.    Do others agree with my explaination?
15050
Prove that $$\frac {1} {2\pi i} \oint \frac {{\rm d}z} {P(z)} $$ over a closed  smooth simple curve that contains all of the roots of the polynomial $P(z)$ is either zero if $n \geq 2$ or equals $\frac{1} {a_0} $ if it's degree is $1$.   Proof: $n\geq 2$ $$\frac {1} {P(z)=(z-z_0)(z-z_1)\dots(z-z_n)}$$ So after partial decomposition of the fraction I'll have something like $$\frac {A} {z-z_0} +\frac{B} {z-z_1}+\cdots$$ where $A,B \in \mathbb{C}$ (Is this fact true that $A,B$ wont be polynomials and just coefficients I can see it why in an intuitive way but cannot write a full proof of it for the decomposition trying to get a linear system or something) then use Cauchy's theorem and integrate all of these on  different circles with their center at each root  and argue that  the integral is the same as the sum of those circle integrals. Now since $\frac {1}{z-z_n}$ will be analytic inside the circle when I'm integrating on circles that do not contain $z_n$  some  of those integrals  will be zero. and now since my curves are of the form $\gamma(t)=z_n+e^{it}$ , $t \in [0,2\pi]$ the rest of the integrals  will look like $$\int_0^{2\pi}  \frac{Aie^{ti}} {e^{it}}\,{\rm d}t $$  and all of them will be $A_i$.   Is my proof correct? Only part I cannot show is that I won't have any problem with decomposing the fraction. And I'm also using the assumption that after the decomposition I'll have as numerators simple complex numbers not not functions of any kind.
51130
Let $f:U\rightarrow U$  be analytic, where $U \subsetneqq \mathbb{C}$ is a simply connected domain. Suppose that $f$ has at least two fixed points. Prove that $f$ is the indentity map.  I am able to prove the theorem in the case where $U = \mathbb{D}$ is the unit disc (the proof uses Schwarz lemma, and is discussed ).  The proof however skips the (rigorous) details of how the Riemann Mapping Theorem allows us to reduce from $U$ to $\mathbb{D}$. Any details are appreciated.
110980
Let $\gamma:I\to\mathbb{R}^2$ be a closed plane curve, for simplicity, a unit circle. Therefore, we have $$\gamma(\varphi) = (\cos \varphi, \sin \varphi).$$ What is the proper domain of $\varphi$?  says it's $\varphi \in [0,2\pi]$ with $\gamma(0) = \gamma(2\pi)$. What is the advantage of this domain rather than a domain $\varphi\in[0,2\pi\rangle$ with the additional condition that $$\lim_{\varphi \to 2\pi}\gamma(\varphi)=\gamma(0)?$$ It seems to me that the second definition is much more natural since no point on the curve is repeating, and the usual angle variable in the polar cordinates uses this domain.
68304
The following problem comes from do Carmo's book Differential Forms and Applications, Chapter 2, Exercise 4:  Let $\omega$ be a differentiable 1-from defined on an open subset $U\subset \mathbb{R}^n$. Assume that for each closed differential curve $C$ in $C$, $\int_C \omega$ is a rational number. Prove that $\omega$ is closed.   Can anyone give some hints? My idea was that we could try to show the integral vanishes if the curve lies in $N_\epsilon (p)$ (i.e. a small neighborhood around $p$) for any $p$ in $U$ and use the Poincare Lemma, but failed.
70820
How can I prove the following?  $\gamma (t)$ is unit speed, $\dot \gamma(t) \not= 0 \Rightarrow \ddot \gamma(t)$ is perpendicular to $\gamma(t)$   I don't really see where a problem would arise when $\dot \gamma(t)=0$ would cause such a problem
61774
$\int_0^\infty \frac{1}{1+x^9} \, dx$  I tried taking the integral of $\Gamma_R = [0,R] \cup \gamma_R \cup I_R$, where we see that \gamma_R is the circle parametrized by $z = Re^{it}$ with $t\in[0,\frac{\pi}{2}]$. And $I_R$ is the line form $iR$ to $0$  one can note the following three things:  1) $\int_0^\infty \frac{1}{1+x^ 9}\,dz$ = $\int_0^\infty \frac{1}{1+z^ 9}\,dz$. (Note that $z$ is on the $x$-axis)  2) $\int_{\gamma_R}\frac{1}{1+z^ 9} = 0$. (By the ML inequality)  the main problem is when i parametrize $I^{-}_{R}$ with $z(x) = xi$ and $x \in [0,R]$. The main problem is that i get that:  3) $\int_{I_R}\frac{1}{1+z^9} = -\int_{I_R^-}\frac{1}{1+z^9} = -\int_{0}^{R} \frac{i}{ix^9 + 1}$.   note that i want to work towards taking $R$ to infinity eventually and then equal the sum of the integrals to the sum of the residuals of $\Gamma_R$ times $2\pi i$. The main problem here is that my term in 3), does not look like the form $c \cdot \int_{0}^{\infty}\frac{1}{1+x^ 9}$ with c a constant value. If i can get this done i think i'm finished.  PS: The singular points i got where $e^{\frac{\pi i}{9}},e^{\frac{3 \pi i}{9}}$.
76647
Assume that $f(z)$ is holomorphic and satisfies the inequality $|f(z)-1|&lt;1$ in a region $|z|&lt;R$. Show that $$\int_\gamma\frac{f'(z)}{f(z)}dz=0$$ for all closed curves $\gamma$ in $|z|&lt;R$. (The continuity of $f'(z)$ is taken for granted.)   The anti-derivative of $\dfrac{f'(z)}{f(z)}$ is just $\ln(f(z))$, which is holomorphic as long as $\ln(f(z))$ avoids the log branch cut. The log branch cut is the negative real axis, $a\leq 0$.   I want to use the fact that: If $f$ is the derivative of a holomorphic function in the region $\omega$, then the integral $\int_\gamma fdz$ depends only on the endpoints of $\gamma$.  To use this, we only need the condition that $f(z)$ is not a non-positive real number. The condition $|f(z)-1|&lt;1$ seems like an overkill. (Of course it does imply that $f(z)$ is not a non-positive real number.) But am I misunderstanding something?