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If $f$ is as given in the problem statement, then how do I determine its image?

My book says that the image of $f$ = {$z$ in the nonzero complex numbers under multiplication such that $z=f(x)$ for some $x$ in the reals under addition}.

SO could I take $x=3$ and then $f(x)=e^{ix}=3$, so image of $f$ equals 3?

This problem is so frustrating.

  • Actually that one is a classical. It would be a good idea to keep it as math culture. – Mauricio Tec Sep 08 '13 at 04:02
  • @AbhishekMallela As stated, I think there is more to the problem than T.Bongers' solution, unless I am missing something? – Evan Sep 08 '13 at 04:52

2 Answers2

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The image of $f$ is precisely the points in $\Bbb{C}^*$ which are hit by $f$; that is, the image is precisely the set $\{f(x) : x \in \Bbb{R}\}$.

So if, for example, $f(x) = e^{ix}$, then the image is the set of points in the complex plane which can be written as $e^{ix}$ for some $x$. More precisely, these are the points with absolute value $1$ and the image of $f$ is $\{z : |z| = 1\}$.

  • Why are these only the points with absolute value 1? Why not $e^{ix}$ = 3, or $e^{ix}$ = 5364? –  Sep 08 '13 at 03:54
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    @AbhishekMallela $|e^{ix}| = 1$ regardless of what $x$ is. Use Euler's identity to get $e^{ix} = \cos{x} + i \sin{x}$ so that $$|e^{ix}|^2 = \cos{x}^2 + \sin{x}^2 = 1$$ –  Sep 08 '13 at 03:55
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You need to use homomorphism properties, i.e. $f(x+y) = f(x)f(y)$. One can show that this functional equation leads to an exponential form (if $f$ is assumed to be continuous). $f(x) = e^{ix}$ is one such function, but there are also others :)

In fact, nobody said $f$ had to be continuous.. I think there are weird examples too?

Evan
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  • [ I believe $f(x) = e^x$ is still a homomorphism, it just keeps real (still nonzero) . Real numbers are complex numbers too ] – Evan Sep 08 '13 at 03:59
  • In fact, to prove that $f$ is an exponential it is sufficient to assume that either a) $f$ is continuous at a single point or b) $f$ is measurable. http://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function

    Using the axiom of choice, you could take a discontinuous additive map $h : \mathbb{R} \rightarrow \mathbb{R}$ and look at the composition $g \circ h$ where $g$ is an exponential map $g(x)=e^{ix}$. That'd be pretty weird! http://math.stackexchange.com/questions/22069/is-there-a-name-for-function-with-the-exponential-property-fxy-fx-times-f

     – dls Sep 08 '13 at 04:56
  • @dls Ah yes, I remember reading this somewhere before, thanks for pointing them out! – Evan Sep 08 '13 at 04:57