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Let $f$ be a non-constant entire function.Which of the following properties is possible for each $z \in \mathbb{C}$

$(1) \ \ \mathrm{Re} f(z) =\mathrm{Im} f(z)$

$(2) \ \ |f(z)|<1$

$(3)\ \ \mathrm{Im} f(z)<0$

$(4)\ \ f(z) \neq 0 $

I tried for $(2)$ and $(3)$ option.For $(2) f$ is entire and bounded by Louiville's theorem it has to be constant which is contradiction to hypothesis.for $(3)$ if imaginary part or real part is bounded below or above then function has to be constant.How eliminate $(1)$ & $(4)$? I don't know what are the right option.

Here it is possible that there are more than one answers. Please help me thanks in advance.

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    $\begingroup$ Think of the non-constant entire functions you know. Which of the above properties do they satisfy? $\endgroup$
    – Ron Gordon
    Commented Jan 24, 2014 at 16:25
  • $\begingroup$ I suppose first thing you should have tried for first case is cauchy riemann... you tried that? $\endgroup$
    – user87543
    Commented Jan 24, 2014 at 16:29

3 Answers 3

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1. Consider the function , $g(z)=e^{f^2(z)}$. Then , $g$ is entire (as $f$ is). Now , $$|g(z)|=\left|e^{(u^2-v^2)+2iuv}\right|=e^{u^2-v^2}=1.$$where , $f=u+iv$ and given $u=v$. So , $g$ is entire and bounded and hence constant. So , $f$ is constant ,which contradicts. So , it is FALSE.

2. Clearly $f$ is bounded and entire and hence constant. So , it is FALSE.

3. Consider , $g(z)=e^{-if(z)}$. Then $g$ is bounded and entire and hence constant and so $f$ is constant. So , it is FALSE.

So, only 4. is TRUE.

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$f(z)=u(x,y)+iu(x,y)$

$u_x=u_y$ and $u_y=-u_x$

Can you conclude now?

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  • $\begingroup$ In the first option it can't be hold. $\endgroup$ Commented Jan 24, 2014 at 16:47
  • $\begingroup$ so... what do you want to say? $\endgroup$
    – user87543
    Commented Jan 24, 2014 at 16:47
  • $\begingroup$ So,first is not true. $\endgroup$ Commented Jan 24, 2014 at 16:50
  • $\begingroup$ How to prove last option by assuming $f(z)=0$ for some $z$ in $\mathbb{C}$ and function is given entire Can i prove that function is zero on complex plane. $\endgroup$ Commented Jan 24, 2014 at 16:56
  • $\begingroup$ Thanks Cameron Buie and Praphulla Koushik both of you. $\endgroup$ Commented Jan 24, 2014 at 17:01
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Hints: For $1,$ note that $f$ can't be $0$ everywhere, nor can its derivative. Hence, there is some non-empty open set that $f$ maps to an open set. (Why?) Can the line $\operatorname{Re}(w)=\operatorname{Im}(w)$ contain any non-empty open set?

For $4,$ try to think of an example of a non-constant entire function that is never $0.$ (A basic example should do.)

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  • $\begingroup$ If I consider $f(z)=e^z$ then $(4)$ holds and there are many entire function. $\endgroup$ Commented Jan 24, 2014 at 16:49
  • $\begingroup$ Bingo! That tells you that $4$ is a possibility. All that's left is for you to rule out option $1.$ You can proceed in the way I mentioned, or alternately, you can use the Cauchy-Riemann equations. $\endgroup$ Commented Jan 24, 2014 at 16:51
  • $\begingroup$ Can i prove last one? If I amssuming $f(z)=0$ for some $z$ and function is entire then can i prove fuction is zero everywhere? $\endgroup$ Commented Jan 24, 2014 at 16:54
  • $\begingroup$ You have already proven the last one. $f(z)=e^z$ is a nonconstant entire function such that $f(z)\ne0$ for all $z.$ Thus, it is possible for a nonconstant entire function to be nonzero everywhere. Proof by example complete. $\endgroup$ Commented Jan 24, 2014 at 16:56
  • $\begingroup$ You cannot prove that if an entire function has a zero, then it is zero everywhere, since it isn't true. Think of some other basic examples of entire functions to see this. $\endgroup$ Commented Jan 24, 2014 at 16:57

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