Let $f(z)=\frac{z^{5}}{|z|^4}$ for ($z\neq 0$) and $f(z)=0$ for $z=0$ then $u$ and $v$ satisfy Cauchy-Riemann equations everywhere but $f(z)$ is not differentiable at $z=0$ which means its not analytical because for a function to be analytic it must be differnetiable everywhere. I am confused because my teacher said that if $u$ and $v$ satisfy Cauchy-Riemann equations then $f(z)$ is analytic but this example proves that this is false. Am I misunderstanding something or is my teacher wrong?
88421
It is, of course, one of the first results in basic complex analysis that a holomorphic function satisfies the Cauchy-Riemann equations when considered as a differentiable two-variable real function. I have always seen the converse as: if $f$ is continuously differentiable as a function from $U \subset \mathbb{R}^2$ to $\mathbb{R}^2$ and satisfies the Cauchy-Riemann equations, then it is holomorphic (see e.g. Stein and Shakarchi, or ). Why is the $C^1$ condition necessary? I don't see where this comes in to the proof below.   Assume that $u(x,y)$ and $v(x,y)$ are continuously differentiable and satisfy the Cauchy-Riemann equations. Let $h=h_1 + h_2i$. Then \begin{equation*} u(x+h_1, y+h_2) - u(x,y) = \frac{\partial u}{\partial x} h_1 + \frac{\partial u}{\partial y}h_2 + o(|h|) \end{equation*} and  \begin{equation*} v(x+h_1, y+h_2) - v(x,y) = \frac{\partial v}{\partial x} h_1 + \frac{\partial v}{\partial y} h_2 + o(|h|). \end{equation*} Multiplying the second equation by $i$ and adding the two together gives  \begin{align*} (u+iv)(z+h)-(u+iv)(z) &amp;= \frac{\partial u}{\partial x} h_1 + i \frac{\partial v}{\partial x} h_1 + \frac{\partial u}{\partial y} h_2 + i \frac{\partial v}{\partial y} h_2 + o(|h|)\\\  &amp;= \left( \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \right) (h_1+i h_2) + o(|h|). \end{align*}  Now dividing by $h$ gives us the desired result.   Does there exist a differentiable but not $C^1$ function $f: U \rightarrow \mathbb{R}^2$ which satisfies the Cauchy-Riemann equations and does NOT correspond to a complex-differentiable function?
46926
$f$ is analytic function on open unit disk, and $|f(z)|\le 1-|z|\forall z\in D$, we need to show $f\equiv 0$, just a hint please.
112409
This is a question from an old prelim in complex analysis. Show that if $f$ is an analytic function such that $|f(z)| \leq \frac{1}{1-|z|}$ for $|z|&lt;1$, then $|f'(0)| \leq 4$. I tried using both Cauchy's $ML$-Estimates along with the Cauchy Integral Formula and even the Schwarz-Pick Lemma, but I still do not get the desired result. Any suggestions as to how best to proceed?
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 ?
117479
Let $f$ be an analytic function on $\bar{𝐷} = \{z \in \mathbb{C}: |z| \le 1\}$. Assume that $|𝑓(𝑧)| ≤ 1$ for each $z\in \bar{D}$. Then, which of the following is NOT a possible value of $(e^{f})''(0)$??  $(A) 2$  $(B) 6$  $(C) \frac{7}{9}e^{\frac{1}{9}}$  $(D) √2 + 𝑖√2$  So I take a look at Cauchy Integral Formula and then $$(e^{f})''(0)=\frac{1}{\pi}\int_{|z|=1}\frac{e^{f(z)}}{z^3}$$  Taking modulus on both sides we have $$|(e^{f})''(0)| \le \frac{e}{\pi}2\pi=2e$$  Hence (b) cannot be the choice.   Is this alright!!  Thanks for the 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 $|g'(0)| \leq 1$.
69895
I am supposed to integrate $f(z)=$$\frac{5}{z}$ from -3 t0 3 but I am having trouble understanding how to do this. I've done the integration the "hard" way by using parametrizations but now I need to use the fact that $f(z)$ is analytic everywhere except at $z=0$ and a different method to perform the integration but I am stumped.
18552
$f$ is an analytical function on the unit disk $\mathbb{D}$ for which there is a $C &gt; 0$ such that  $\forall z \in \mathbb{D} \colon |f(z)| \leq \dfrac{C}{1-|z|}$.  I know have to prove that    $\forall z \in \mathbb{D} \colon |f'(z)| \leq \dfrac{4C}{(1-|z|)^2}$.  I tried using the Cauchy integral formula but I can't find a fitting circle to get this result.
66397
Use the Cauchy-Riemann equations to show that the function $$g(z) = \sin (\bar z)$$ is not analytic at any point of $\mathbb{C}$.  Here's as far as I got -  $$\sin \left(\frac{\bar z}{1}.\frac{z}{z}\right) =\sin \left(\frac{|z|^2}{z}\right) =\sin \left(\frac{x^2 + y^2}{x+iy}\right)$$  I can't see how to separate the real and imaginary parts so that I can apply the Cauchy-Riemann equations.
94221
I'm reading through a development of the maximum modulous principle, but I am stuck verifying a remark:  $$\text{"it is enough to show that $|f|$ is constant, from which we may conclude that $f$ is.''}$$  So I am trying to prove it as a lemma:  Let $f$ be analytic on a domain $D$.  If $|f|$ is constant then so is $f$.  I tried using the fact that $|f|$ constant implies $|f|$ is analytic.  From here this means that $Re(|f(x,y)|)$ is harmonic.  I wrote out the consequence to this using Laplace's Equation, hoping to force the partial derivatives of $f$ to vanish, but it didn't seem to go anywhere.  Any suggestions?
86184
Let $G\subseteq\mathbb{C}$ be a domain and $f\colon G\to\mathbb{C}$ holomorphic. Show:     $$ f'=0\mbox{ in }G\Rightarrow f\mbox{ is constant} $$      Cauchy-Riemann: $$ f'(z)=f_x(z)=u_x(x,y)+i v_x(x,y), z=x+iy=0\\ \Leftrightarrow u_x(x,y)=0~\wedge v_x(x,y)=0 $$  and $$ f'(z)=-if_y(z)=-iu_y(x,y)+v_y(x,y)=0\\\Leftrightarrow u_y(x,y)=0~\wedge~v_y(x,y)=0 $$  That means: If one derivates u and v to x, one gets 0, and therfore the x-part of these functions must be constant. The ssame for the y-part of u and v. So u and v are constant and so is then f.  Is that okay?
93399
I've been thinking how to prove that an analytic function $f$ is a constant if the absolute value of $f$ is a constant, but I haven't figured it out yet.  What I was thinking is to use Cauchy-Riemann equations, but it didn't work well...  If this is not true, I would like to know the counterexample...  Here is what I tried:  $$|f|=|u+iv|=\sqrt {u^2+v^2}$$  Thus $u^2+v^2$ is a constant.  (1)    $\displaystyle u\frac {\delta u}{\delta x}+v\frac {\delta v}{\delta x}=0 $  (2)    $\displaystyle u\frac {\delta u}{\delta y}+v\frac {\delta v}{\delta y}=0 $  Plug Cauchy Riemann into (2).  $$\displaystyle -u\frac {\delta v}{\delta x}+v\frac {\delta u}{\delta x}=0 $$  and I'm stuck here...
62466
I'm talking about complex function   $f$ is analytic function on a region $D$ that include the point $z=0$. for every $n\in N$ such $\frac{1}{n}$ is in $D$ , the function follows this condition $|f(\frac{1}{n})| \leq \dfrac{1}{2^n}$   is that function must be constant ?
111761
Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be an holomorphic function such that $f'(z_0) \neq 0$ for some $z_0 \in \mathbb{C}$.Prove that there is $r&gt;0$ such that, if $|z-z_0|&lt;r$ and $z \neq z_0 $, then $f(z) \neq f(z_0)$.  It seems very trivial, but I can't solve it...
91292
Let $A$ be an open connected subset of $\mathbb C$ and $f:A\to B$ analytic. Suppose there exist reals constants $a$, $b$ and $c$ such that $a^2+b^2\neq 0$ and $a\operatorname{Re}\bigl(f(z)\bigr)+b\operatorname{Im}\bigl(f(z)\bigr)=c$ for all $z \in \mathbb C$. Prove $f$ is constant.  Proof: as f is analytic then cauchy-riemann equations holds.  I computed the partial derivatives of $a\operatorname{Re}\bigl(f(z)\bigr)+b\operatorname{Im}\bigl(f(z)\bigr)=c$ with respect to $x$ and $y$ and tried to make a relation with the cauchy equations but things become complicated, because I got this:  $a\frac{\partial u}{\partial x}=a\frac{\partial v}{\partial y}=-b\frac{\partial v}{\partial x}=b\frac{\partial u}{\partial y}$ and I don't know how to get the result.
60820
Is there a characterization of the nonconstant entire functions $f$ that satisfy $|f(z)|=1$ for all $|z|=1$?   Clearly, $f(z)=z^n$ works for all $n$. Also, it's not difficult to show that if $f$ is such an entire function, then $f$ must vanish somewhere inside the unit disk. What else can be said about those functions?  Thank you.
48119
I want to prove that if $f$ is an analytic function for which $\Im f$ is constant, this implies that $f$ itself is constant.  So to start off, it's not given that the function is entire or anything, ruling out Liouville. Rather, I'm guessing the maximum modulus theorem will prove useful here (?).  If I were to show using that the real part of $f$ was constant, proving the statement would be easy (just take the absolute value and investigate $e^f$), but in this case I'm none the wiser from that. Should I use the same parametrization $e^f$ and apply the Cauchy-Riemann equations somewhere? Any suggestions welcome.