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I was thinking about the problem that says:

Let $f$ be an entire function whose values lie in a straight line in the complex plane. Then which of the following option(s) is/are correct?

(a) $f$ is necessarily identically equal to zero,

(b) $f$ is constant,

(c) $f$ is Möbius map,

(d) $f$ is a linear function.

From the fact that the values of f lie in a straight line,we can take f(x) to be of the form ax+b; a,b being constants. But in this case ,we can not conclude that f is a linear function as the term linear function refers to a function that satisfies the following two properties:

f(x+y)=f(x)+f(y) and f(ax)=af(x). But we see that f does not satisfy the very first property. So option (d) does not hold good.We can also eliminate (a),(b) in the process.So i think $f$ represents a Möbius map.Does my thinking go in the right direction?

Please help. Thanks in advance for your time.

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What have you tried? –  user27126 Dec 8 '12 at 5:59
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"From the fact that the values of f lie in a straight line, we can take f(x) to be of the form ax+b." I'm not so sure about this... The situation you have is a holomorphic function $w = f(z)$ whose image is a line. You seem to be confusing this with a real-valued function of a real variable $y = f(x)$ whose graph is a line. These are very different things. –  Jesse Madnick Dec 8 '12 at 7:39

2 Answers 2

up vote 7 down vote accepted

You can use the Open Mapping Theorem, which states:

Any non-constant holomorphic function that is defined on a non-empty open subset $ U $ of $ \mathbb{C} $ must map open subsets of $ U $ to open subsets of $ \mathbb{C} $.

As $ f[\mathbb{C}] $ is forced to be contained in a straight line, and as a straight line cannot contain a non-empty open subset of $ \mathbb{C} $, it follows that $ f $ must be a constant function. The correct answer is therefore (b).

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Did you really mean (d)? –  robjohn Dec 8 '12 at 7:34
    
Thanks for spotting that typo error! –  Haskell Curry Dec 8 '12 at 7:43
    
@HaskellCurry"straight line is not an open subset of C"...can you explain a little bit more,sir? " The correct answer is therefore (b)"...Is it (b) or (d), i am confused from robjohn's comment. –  learner Dec 8 '12 at 7:54
    
@HaskellCurry Thank you sir for the clarificaton. –  learner Dec 8 '12 at 7:58
    
The correct answer is (b). I initially typed (d). –  Haskell Curry Dec 8 '12 at 8:06

This is really an extended comment. You seem to have a lot of misconceptions about linear functions. Let me clarify some things:

(1) As I said in the comments: the situation you have is a complex-valued function of a complex variable $w = f(z)$ whose image is a line. You cannot conclude from this alone that $f(z) = az + b$ for some constants $a, b$.

I think you are getting confused with the case of a real-valued function of a real variable $y = f(x)$ whose graph is a line. In this case you can say that $f(x) = ax + b$.

There are two distinctions here. First, there is a difference between real functions of a real variable and complex functions of a complex variable. Second, there is a difference between "image" and "graph."

(2) You have to be careful when you say "linear function."

In many calculus classes -- and unfortunately also in some analysis classes -- the term "linear function" means a function of the form $f(x) = ax + b$.

However, this is not the best terminology. A function of the form $f(x) = ax + b$ should be called an affine function. A linear function is, as you say, one that satisfies $f(x+y) = f(x) + f(y)$ and $f(kx) = kf(x)$.

From this, one can show that (for real-valued functions of a real variable), the true linear functions are of the form $f(x) = ax$. That is, linear functions are a special kind of affine function (the ones with $b = 0$).

Is all of this clear?

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thanks a lot sir, for the clarification.I have got it,now. –  learner Dec 8 '12 at 9:31

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