Let $f = u +iv$ be holomorphic in domain $D$ such that $f$ is nonezero in $D$. prove that $$ h=\frac{u \frac{\partial u}{\partial x}+v \frac{\partial v}{\partial x}}{u^{2}+v^{2}} $$ is harmonic I tried to apply Cauchy-Riemann equations and didn't find anything, what is the &quot;trick&quot; here? what is the right attitude?
120456
We have an open set $U$ in $\mathbb{R}^2$ and two functions $u, v \in C^1 (U)$. $u$ and $v$ satisfy the Cauchy-Riemann equations in $U$, i.e., $u_x = v_y$ and $u_y = -v_x$. Also, $u^2 + v^2 \neq 0$ for all $(x,y) \in U$. Then, we have to show that $$\frac{uu_x + vv_x}{u^2 + v^2}$$ is harmonic in $U$, i.e., the Laplacian of this function is zero.  On the face of it, it seems there is nothing to do but calculate the second partial derivatives and verify the result. But that would be very messy and complicated. Is there something easier that I am missing?  Note: $u_x$ means the partial derivative of $u$ with respect to $x$, etc.
229
I am completely stuck in the following problem:  let f be holomorphic function on the unit disk $\{z:|z|\lt 1\}$ in the complex plane.which of the following is/are necessarily true?  (a) if for each positive integer $n$ we have $f(\frac{1}{n})=\frac{1}{n^2}$,then $f(z)=z^2$ on the unit disc.  (b) if for each positive integer $n$ we have $f(1-\frac{1}{n})=(1-\frac{1}{n})^2$,then $f(z)=z^2$ on the unit disc.  (c) $f$ cannot satisfy $f(\frac{1}{n})=\frac{(-1)^n}{n}$ for each positive integer $n$  (d) $f$ cannot satisfy $f(\frac{1}{n})=\frac{1}{n+1}$ for each positive integer n   what I thought in this problem is the following:  in the first part,by identity theorem we can say that $f(z)=z^2,\forall$z in the unit disc.  but,in the second part ,if we choose $g(z)=z^2$ then,$f(1-\frac{1}{n})=g(1-\frac{1}{n})$ but,the limit point ,i.e, $1$ is not in the unit disc.so,in this case,we cannot use identity theorem to conclude whether $f(z)=z^2$ on the unit disk or not .so,what procedure can we follow in this case???also,explain the options (c) and (d).  please answer in detail..  thanks a lot...
18107
I'm trying to do this old qualifying exam problem from UW Madison. Let $D^\ast=\{z\in\mathbb{C},0&lt;|z|&lt;1\}$ and $f$ be a non constant holomorphic function on $D^\ast$. Assume that $\text{Im} f(z)\geq 0$ if $\text{Im} z\geq 0$ and $\text{Im} f(z)\leq 0$ if $\text{Im} z\leq 0$. Prove that if $z\in D^\ast$ is not real, then $f(z)$ is not real. Show that if $z\in (-1,0)\cup(0,1)$, then $f'(z)\not=0$. Prove that $0$ is either a removable singularity with $f'(0)\not=0$ or $0$ is a simple pole of $f$. What I have thought of so far: If $z\in D^\ast$ and $\text{Im} z&gt;0$, but $\text{Im} f(z)=0$, then apply the maximum modulus principle on $\{z|z\in D^\ast,\text{Im} z&gt;0\}$ to $e^{if}$ to obtain a contradiction. This shows that if $z\in D^\ast$ and $\text{Im} z&gt;0$, then $\text{Im} f(z)&gt;0$. Similarly if $z\in D^\ast$ and $\text{Im} z&lt;0$. Furthermore, the reflection principle shows that $f(\overline{z})=\overline{f(z)}$. If $z\in(-1,0)\cup(1,0)$, to show that $f'(z)\not=0$, we use the following fact: fact: If $f$ is a complex valued function continuous on $\overline{D(0,R)}$ and holomorphic on $D(0,R)$, then for any $z\in D(0,R)$, we have $$f(z)=\int_0^{2\pi}i \text{Im} f(\xi)\frac{\xi+z}{\xi-z}\frac{d\theta}{2\pi}+K$$ for some constant $K$, where $\xi=Re^{i\theta}$. We can differentiate the expression to get an expression of $f'(z)$ in terms of $\text{Im} f$. This same fact shows that $f'(0)\not=0$ if $0$ is a removable singularity. If $0$ is a pole, $\text{Im} z$ is dominated by the imaginary part of $\frac{C}{z^n}$ for some $n\in\mathbb{Z}^+$ and $C\in\mathbb{R}$ when $|z|&gt;0$ is small, then unless $n\not=1$, we can find some $z\in D^\ast$, $\text{Im} z&gt;0$ such that $\text{Im} f(z)&lt;0$. So if $0$ is a pole, then it is a simple pole. My question is: how to show that $0$ is not an essential singularity? Thanks!!
1243
Find all holomorphic functions $f:\mathbb C\to\mathbb C$ such that for all $x,y\in\mathbb R$ $\Re(f(x+iy))=2xy$  So $f$ must be complex differentiable, which is equivalent to that Cauchy-Riemann equations must be satisfied Let $f(x+iy)=P(x,y)+iQ(x,y)$ (Here $P(x,y)=2xy$) then $\frac{\partial P}{\partial x}=2y=\frac{\partial Q}{\partial y}$ $\implies Q(x,y)=y^2+Q(x)$ the second equation is: $\frac{\partial P}{\partial y}=2x=-\frac{\partial Q}{\partial x}$ $\implies Q(x)=-y^2$ Hence $Q(x,y)=x^2-y^2\implies f(x+iy)=2xy+i(x^2-y^2)$ Can you verify my steps, and holomorphic means differentiability in the complex sense right ?
76926
Holomorphic functions are one of the most beautiful objects in mathematics. However, Cauchy-Riemann equations are a little bit of a mystery for me. What does that mean intuitively when a function satisfyes those equations?
122647
The problem is:  Suppose that $f(z)$ is continuous on a domain $D$ that contains the unit circle, and that $f(z)$ satisfies: $$\vert f(e^{i\theta})\vert \leq M \; \forall \theta \, \in [0, 2\pi )$$ and $$\bigg \vert \int _{\vert z \vert=1} f(z) \text{d}z \bigg \vert =2\pi M $$ show that there exists a $c\in \mathbb{C}$ such that $f(z)=c \bar{z}$ for all $\vert z\vert=1$.  I've tried lots of different approaches such as defining a function $g(z)=\frac{f(z)}{\bar{z}}$ and trying to show that $g'(z)=0$ so that $g$ is constant on the circle, or trying to use some bounding theorems for analytic functions to show that $g$ is constant but this requires $g$ to be analytic. Some other methods were using the Cauchy-Riemann equations but I couldn't find any reason why $g$ should be analytic.  Any help would be greatly appreciated, thanks!
106718
Let $f$ be a continuous function on the closed unit disk and holomorphic inside it, such that $f(z) = 0$ for all $ z = \exp (it) $ with $0 ≤ t ≤ \pi/2 $. Prove that $ f(z) = 0 $ for all $ z \in \overline{D}(0,1) $.  I have worked this exercise for absurdity, but it has not reached any conclusion. I would like to know a more sophisticated test, perhaps by a direct method.
82525
Prove that the functions $f(z)$ and $\overline{f(\overline{z})}$ are simultaneously holomorphic.   I take this to mean that $f(z)$ is holomorphic if and only if $\overline{f(\overline{z})}$ is holomorphic.  Let $g(z)=\overline{f(\overline{z})}$. Note that $\overline{g(\overline{z})}=f(z)$. So it suffices to prove that if $f(z)$ is holomorphic, then $g(z)$ is holomorphic.  Write $f(z)=u(z)+iv(z)$. Since $f(z)$ is holomorphic, the real and imaginary parts satisfy the Cauchy-Riemann equations:  $$\frac{\partial{u(z)}}{\partial{x}} = \frac{\partial{v(z)}}{\partial{y}}, \frac{\partial{u(z)}}{\partial{y}} = -\frac{\partial{v(z)}}{\partial{x}}.$$  We have $g(z) = u(\overline{z})+i(-v(\overline{z}))$. To prove that $g(z)$ is holomorphic, we must prove that its real and imaginary parts satisfy the Cauchy-Riemann equations:  $$\frac{\partial{u(\overline{z})}}{\partial{x}} = \frac{\partial{(-v(\overline{z}))}}{\partial{y}}, \frac{\partial{u(\overline{z})}}{\partial{y}} = -\frac{\partial{(-v(\overline{z}))}}{\partial{x}}.$$  How can we obtain this from the above relations?
8166
This is from an old qualifying examination question.  If f is holomorphic in the unit disk $D$ and $|f(z)|&lt;1$ for all $z\in D$. Suppose also that $f$ has two distinct fixed points in $D$ then $f(z)=z$ for every $z\in D$  I know that I have to use the Schwarz's lemma and may be make use of Möbius transformations. I tried setting $g(z)=\phi_a\circ f\circ \phi_{-a}$. But that does not seem to work because I don't see why $|g(z)|=|z|$ for some non zero $z$.   Any helpful hints are greatly appreciated.  Edit: Of course the $a$ above is one of the fixed points.
29606
While studying for an exam in complex analysis, I came across this problem. Unfortunately I was not able to solve it. Any help would be greatly appreciated.  Let $U ⊂ \mathbb{C}$ be a domain and $f_1, . . . , f_n : U \rightarrow \mathbb{C}$ holomorphic functions, such that $\sum_{k=1}^n \left| f_k(z) \right|$ is constant on $U$. Show that $f_1, . . . , f_n$ are constant.
114483
This is one of the past qualifying exam problems that I was working on.   I know that, when we let $z=x+iy$, ${|z|}^2=x^2+y^2$ is not harmonic. I do not know where to start to prove that there is no harmonic function that is positive everywhere.  Any help or ideas idea will be really appreciated.   Thank you in advance.
117438
Let $u:\mathbb{C}\rightarrow\mathbb{R}$ be a harmonic function such that $0\leq u(z)$ for all $ z \in \mathbb{C}$. Prove that $u$ is constant.  I think i should use Liouville's theorem, but how can i do that? Help!
18801
Suppose $f(z)$ is holomorphic in $|z|&lt; 1$ and $\operatorname{Re}f(z) &gt; 0$ in this region. Furthermore, suppose $f(0) = a &gt; 0$. Then why does $|f'(0)| \leq 2a$?  The hint I have is to consider $g(z) = \frac{f(z) - a}{f(z) + a}$ and then show $|g'(0)| \leq 1$. If I do have $|g'(0)| \leq 1$, then the result follows as $|f'(0)| = 2a|g'(0)| \leq 2a$, but I have no idea on how to show $|g'(0)| \leq 1$.
63589
"Let $u, v$ be harmonic functions on a region G. Prove that if the product $uv$ is identically zero, then either $u$ or $v$ must be identically zero." Could sb. give me a hint for this question? I tried all the properties and theorems on harmonic functions (definition, poisson integral, max. principle, passed a holomorphic function,..), but could go nowhere.  Thanks.
122566
Let $f$ be holomorphic non-constant on $D=\left\{ 0&lt;\left|z\right|&lt;10\right\}$ . Given that for all $n\in\mathbb{N}$: $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ prove that $f$ has an essential singularity at $0$. Find an example of $f$ satisfying the condition.   My idea was to assume for contradiction the singularity is not essential, which means that either the limit $z\to0$ of $f$ exists or of $1/f$, and from $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ we get that the limit of $f$ exists and is $0$.  I then define $h(z)=f(z)$ for $z\neq0$ and $h(0)=0$, which is holomorphic on the entire disk, and derive a contradiction from there and the uniqueness theorem. But then the suitable holomorphic function is the Gamma function, which we have not really discussed in class and I'm not sure about it's properties.  Any ways of solving this which avoids using the gamma function?
3814
The problem as stated in R. Remmert's Theory of Complex Functions p. 62., is this:  Let $f = u + iv$ be holomorphic in the region $G \subseteq \mathbb{C}$ and satisfy $u = h \circ v$ for some differentiable function $h : \mathbb{R} \to \mathbb{R}$. Show that $f$ is constant.  I've attacked this computationally in each way I can think of. I've tried to show that $\frac{\partial f}{\partial z}$ is zero; played with the Laplace equation, knowing that $h \circ v$ and $v$ must be hamonically conjugate, etc. I even considered spinning an argument based on the maximum principle.  It seems obvious, but I have not succeeded in a proof yet. Perhaps I am missing a concept?
71420
Suppose $f$ is holomorphic in the unit disk $D=\{z: |z|&lt;1\}$ and $|f(z)|\leq |f(z^2)|$ on $D$. Show that $f$ is a constant function.   My Try:  I wanted to use Strong Maximum Principle, but failed to prove that $f$ is bounded on $D$. I want to try this problem in my own. Can anybody please give me just a hint?