Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be holomorphic on the form $f(x+iy)=u(x)+iv(y)$ where $u,v$ are real functions. Prove that $$f(z)=\lambda z+c$$   for  $\lambda \in \mathbb{R}$ and $c \in \mathbb{C}$.    How do I prove this? I think I have to use the Cauchy-Riemann equations somehow but I can't quite figure it out. Thank you!
122611
Finding all differentiable $f(z) = u(x) + iv(y)$ in $\mathbb{C}$ where $u(x),v(y)$ are real valued functions.   I’m not sure what to do. Would $f$ be differentiable simply if and only if both $u$ and $v$ were differentiable?   My thinking is that if the limit $$ \lim_{h \to 0}\frac{f(z+h)-f(z)}{h} $$  exists then it would be equal to  $$ \lim_{h \to 0}\frac{u(x+h)-u(x)}{h} + i\lim_{h \to 0}\frac{v(y+h)-v(y)}{h} $$  and therefore would exist if they existed. But that doesn’t seem quite right.  I also thought about using Cauchy-Riemann equations, but seeing that I’m trying to find ones that are differentiable rather than those that aren’t, I thought they wouldn’t be of much help.
25402
I am trying to solve the following problem:     Let $f:U\to V$ be a holomorphic function such that $f'(z)\neq 0$ for all $z\in U$. Show that for all $z_0\in U$, there exists a disc $D_\varepsilon(z_0)\subseteq U$ such that $f:D_\varepsilon(z_0)\to f(D_\varepsilon(z_0))$ is bijective.   Atempt:  I am trying to use Rouche's Theorem, but I am stuck on a particular step. Let $z_0\in U$. Since $f'(z_0)\neq 0$, there is a disc $D_r(z_0)\subseteq U$ such that $$f(z)=f(z_0)+(z-z_0)h(z),\quad\forall z\in D_r(z_0)$$ where $h$ is holomorphic on $D_r(z_0)$ and $h(z)\neq 0$ for all $z\in D_r(z_0)$.   Let $0&lt;\varepsilon&lt;r$ to be defined later and fix $w\in D_\varepsilon(z_0)$. To show that $f$ is injective in $D_\varepsilon(z_0)$ is to show that the function $f(z)-f(w)$ has exactly one zero in $D_\varepsilon(z_0)$. But $$F(z):=(z-z_0)h(z)$$ has exactly one zero in $D_\varepsilon(z_0)$, so we might want to apply Rouche's Theorem with $F$ and $$G(z):=f(z)-f(w)-(z-z_0)h(z)=f(z_0)-f(w)=-(w-z_0)h(w)$$ to conclude that $F$ and $F+G$ have the same number of zeros and hence $f$ is injective. But that means that we need to find $0&lt;\varepsilon&lt;r$ such that $|G(z)|&lt;|F(z)|$ on $\partial D_\varepsilon(z_0)$. That is, $$|w-z_0||h(w)|&lt;|z-z_0||h(z)|$$ for all $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$.  Can we find such $\varepsilon&gt;0$?  It seems intuitive since if we expand $h$ in a power series at $z_0$, $h(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$, then $$|w-z_0||h(w)| = |w-z_0|\sum_{n=0}^\infty a_n|w-z_0|^n &lt; |z-z_0|\sum_{n=0}^\infty a_n|z-z_0|^n.$$ for $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$. But this is of course not a proof.
70748
Suppose $h(z)$ is a complex function. I have noticed that $h(z) = f(z)\cdot g(z)$, where $f(z)$ and $g(z)$ are non-holomorphic. Can $h(z)$ be holomorphic?   Can a similar statement be made, if $f(z)$ is holomorphic, but $g(z)$ is not?
111707
Suppose $f$ is a function such that $f:\mathbb C \to \mathbb C$ such that $f(z)f(iz)=z^2$, then we have to show that $\forall z\in \mathbb C f(z)+f(-z)=0 $.   Actually I have solved this problem and have no doubt about it but actually this question came in MADHAVA-2020. So I posted it on this site.   I am not answering the question and inviting all users to answer it.  It is a nice problem.   I will answer it within a week, but I am letting everyone try.
84947
Suppose that $f$ is holomorphic in $\Omega$, and let $g : \Omega → &gt; \mathbb{C}$ be given by $g(z) = \overline{f(z)}$. Prove that $g$ is   complex diﬀerentiable at $a$ if and only if $f'(a) = 0$.     I have tried the basic limit definition for derivatives but couldn't come up with a proper solution.  Also I know that for a complex valued function $f=u+iv$, If $u$ and $v$ have continuous first derivatives, then $f$ should be analytic on $\Omega$. But I'm not exactly sure whether it will be useful here.   Hep would be appreciated
8552
I wonder if this is correct: there is a holomorphic function on an open connected subset $G$ of $\mathbb{C}$ which maps $G$ onto a subset of a straight line, and I have to show that the function is constant.   I thought I can suppose that the straight line is the real axis (otherwise I can find a rotation and a traslation that will do so) and so using the Cauchy-Riemann equations I find that the function is constant since its imaginary part is zero. Is that correct? Thank you
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...
44851
Suppose $f$ is a continuous complex valued function on a domain $\Omega$. Suppose $f^2$ and $f^3$ are holomorphic in $\Omega$. Show that $f$ is also holomorphic in $\Omega$.   Assume $f=u+iv$. I see that if $u,v$ are in $C^1$ then $f^2$ is holomorphic can imply $f$ is holomorphic considering Cauchy-Riemann equations. But I don't know how to get $u,v$ are in $C^1$ by the adding condition $f^3$ is holomorphic. (I can't get $u,v$ from some combinations of real and imaginary part of $f^2$ and $f^3$). Do someone know how to do this?
75175
Let $\Omega$ be a bounded connected open subset of $\mathbb{C}$ containing $0$. Let $f: \Omega \rightarrow \Omega$ be holomorphic and $f(0) = 0$, $f'(0) = 1$. The problem I am working on is to show that $f(z) = z$.  If $\Omega = \mathbb{D}$, then this follows from the Schwarz Lemma. I also know of a solution () which involves looking at the power series coefficients of $f^n := f\circ f \circ f \circ \cdots \circ f$ ($n$ times) and using the Cauchy estimates, but is there a different way of doing this problem that doesn't involve taking the $k$th derivative of $f^n$?
43808
There doesnt exist a holomorphic $f: \mathbb C\setminus0 \rightarrow \mathbb C\setminus 0 $, so that $e^{f(z)} = z $ for all $z \in \mathbb C \setminus 0 $  I tried to work with the derivative of f(z) but i didnt manage to get anything out of it.
94122
I have to solve this problem but i have no idea how to do ...Thanks!  Let be $f:\mathbb{C}\rightarrow \mathbb{C}$ a holomorphic function. Show that if $f\circ f=f$ then f is either constant or $f(z)=z \ \forall\ z\in\mathbb{C}$
92203
Function $f(z)$ is holomorphic at least within the unit disk.   If for all arguments $|z|&lt;1$ we have $|f(z)| \le 1$, and there exists a zero of order $n$ at $z=z_0$, then show that $|f(0)| \le |z_0|^n$.  How should I tackle this problem? Can it be generalized if for $|z|&lt;R$ we have $|f(z)| \le M$ ?    So far I only understand that there exists a holomorphic $g(z)$ with $g(z_0) \ne 0$ such that $f(z) = (z-z_0)^n g(z)$,then $|f(z)| = |z-z_0|^n |g(z)| \le 1$ and moreover $|f(0)| = |z_0|^n |g(0)|$, but that doesn't seem useful since apparently I know too little about $g(z)$. In particular why should $|g(0)| \le 1$?
62345
I came across this question whilst doing some research into complex analysis, and I just can't see what to do!  Let $f(z)$ be a holomorphic function on $\mathbb{C}$. Show that $\overline{f(\overline{z})}$ is holomorphic, whilst $f(\overline{z})$ is holomorphic if and only if $f(z)$ is constant.  I know that holomorphic means that the function is differentiable everywhere, and I need to apply the Cauchy-Riemann equations somehow, but I'm not sure how to approach this.
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?
49561
Is it true that  $ f(\bar{z})=\overline {f(z)}$, Where z is complex?  I think it holds when $f(z)$ is holomorphic since we have $f(z)=p(x,y)+iq(x,y)=p(z,0)+iq(z,0)$ Any help...
32672
Suppose that $f$ is analytic on some region $R\in\mathbb{C}$. If Im$(f)$ = $k\cdot$Re$(f)$ for some nonzero constant $k\in\mathbb{C}$, then show that $f$ is constant on $R$.  I know that if $f'(z)=0$ for all $z$ in some region $R$, then $f(z)$ is constant. However, I'm not sure how to apply this to the question at hand. I also think the Cauchy-Riemann equations could be helpful, but again, I'm not sure how to apply them to the question.  Any guidance would be greatly appreciated. Thank you!
46389
Let $f$ be a holomorphic function defined on $\Bbb C$ such that $f^{-1}(\Bbb R)=\Bbb R$. Prove that there exists $a,b \in \Bbb R, \; a \neq 0 \;$ such that $f(z)=az+b \;$ for every $z \in \Bbb C$.  I was given a hint: define the function $g(z)=\frac{f(z)-f(0)}{z}$. such $g$ is also holomorphic in $\Bbb C$ since the only singularity is in $0$, and it is a removable singularity since we can define $g(0)=f'(0)$. Next I thought about proving $g$ is bounded, and thus constant (using Liouville's theorem), but pretty much out of ideas. How can I finish the proof/solve in another way?