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I was just reading about the e-function in the complex plane and was trying to understand the differences between the real and the complex case. Part of the problem is that the mapping of a 2-D plane to another 2-D plane is hard to visualize.

My question
What are the properties of the complex exponential function in terms of being injective, surjective and/or bijective - and how is this different from the real case? How can you proof these attributes in the real and in the complex case?

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Have you tried working this out yourself? Can you compute e^{x+iy} where x, y are real? (Computing the complex exponential is more or less equivalent to using polar coordinates.) –  Qiaochu Yuan Feb 9 '11 at 15:45
    
@Qiaochu: I always need to visualize these things and, as I wrote, here I have difficulties. But perhaps you can give me a hint howto start differently? –  vonjd Feb 9 '11 at 15:50
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This is more of a book recommendation than an actual hint, but if you're interested in visualizing concepts from complex analysis, take a look at Tristan Needham's Visual Complex Analysis (fitting title, eh?). It's probably my favorite book about complex analysis. –  Jose L. Lykón Feb 9 '11 at 16:00
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I agree with all the recommendations for Needham's book; it's really great. There is also another visualization technique that's not in his book, where one colors each point $z$ in the domain of definition based on the value of $f(z)$ (according to some chosen coloring scheme). I made some images several years ago. The exponential function looks rather dull compared to many other functions, but at least the image clearly shows the periodicity, so here it is anyway: mai.liu.se/~halun/complex/domain_coloring-unicode.html#fig:exp –  Hans Lundmark Feb 9 '11 at 20:39

3 Answers 3

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The complex exponential is never zero, so it's not surjective; this is like the real case. It is also periodic, so it's not injective either; this is unlike the real case.

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The complex exponential $e^z : \mathbb{C} \to \mathbb{C}$ is neither injective nor surjective. The former is because $e^{z + 2 \pi i k} = e^z \forall k \in \mathbb{Z}$ and the latter is because $e^z \neq 0$. In fact, if $e^z = e^w$ then $e^{z-w} = 1 \Rightarrow z - w \in 2 \pi i \mathbb{Z}$, so this is the only source of non-injectivity.

Geometrically, $e^{x + iy} = e^x (\cos y + i \sin y)$ is just the point $(e^x, y)$ in polar coordinates. One can think of the complex exponential as specifying a conformal isomorphism between the following two Riemann surfaces:

  • Its domain mod $2 \pi i \mathbb{Z}$. One should think of this as the complex plane cut up into horizontal strips $S_k = \{ x + iy : 2 \pi k \le y < 2 \pi (k+1) \}$, all of which are then identified, and all of which have their top and bottom borders identified. Roughly speaking this is an infinitely long tube open at both ends.
  • Its range $\mathbb{C} \setminus \{ 0 \}$.

Both of these Riemann surfaces can be identified with the Riemann sphere minus two points. In the first picture one first "puffs out" the tube and then shrinks the points. In the second picture one first identifies $\mathbb{C}$ with the Riemann sphere minus a point via stereographic projection and then removes an additional point.

(This picture works fairly well, I think. For example vertical lines in the domain are mapped to circles in the range in a natural way, and shifting the lines left or right corresponds to varying the radius of the circle, or varying its relative position between the poles on the Riemann sphere.)

There is a nice discussion about how to visualize holomorphic functions in Needham's Visual Complex Analysis.

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Richard Palais has a wonderful program for visualizing some complex maps called 3D-XplorMath which can be downloaded for free at http://3d-xplormath.org/. It also has some nice visual for 3d surfaces and fractals, among other things. This should at least be able to help you visualize exponentiation in $\mathbb{C}$.

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