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How would one find the fixed points of $e^z$, where $z$ is complex (if there are any)? I feel this problem probably has a really obvious answer, and for some reason, I'm just not getting it. Thanks.

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You could start with the definition of e^z when z=x+iy. –  Did Jan 5 '12 at 21:12
    
Here's a bit more...$e^{x+iy}=e^x(\cos(y)+i\sin(y))$ so $|e^{x+iy}|=e^x$. So if $e^{x+iy}=x+iy$ you'll need to have $e^x=|x+iy|=x^2+y^2$ (among other things). –  Bill Cook Jan 5 '12 at 21:28
    
Ok. I would write e^z = e^x * e^iy = e^x * (cos(y) + isin(y))... but then how would I proceed. I've tried manipulating it and it doesn't seem to work. Also, I realised that I need sqrt(x^2 + y^2) = e^x, and y = arctan(y/x).. –  nigelvr Jan 5 '12 at 21:28
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2 Answers

As Martin Brandenburg pointed out on MO, the first value is $-W\left(-1\right) \approx 0.31813 + 1.33723{\rm{i}} \,$ where $W$ is the Lambert W function, which is the principal branch of the inverse function of $f(z) = z e^z.$ So, first, what do we get for $w = W(-1)?$ Well, then $w e^w = -1,$ or $e^w = \frac{-1}{w}.$ If we then take $z = -w,$ we get $$ e^z = e^{-w} = -w = z.$$ So that works.

Next, if $z$ is a fixed point, so is $\bar{z},$ so we get $0.31813 - 1.33723{\rm{i}} $

Finally, sticking with positive imaginary part, we can get a new fixed point by adding, to any existing fixed point $z_n,$ some $\delta_n + (2 \pi + \epsilon_n)\, i$ where $\delta_n > 0$ to accommodate the slightly larger modulus, and $\epsilon_n > 0$ to accommodate the slightly larger argument (approaching $\pi/2$). So $z_{n+1} = z_n + \delta_n + (2 \pi + \epsilon_n)\, i,$ and there should be a sequence of fixed points with positive imaginary part going gently to infinity.

Note that Newton's method does work in the complex plane, it is just that you need to already be very sure you are close. In this case, we can do that, and the error terms $\delta_n, \epsilon_n$ will shrink as $n$ increases.

I've had some time to think about it, we have the nice list by GEdgar for the first few values. In the long run, a good approximation to a fixed point is $$ \approx \log \left( \frac{(4n+1)\pi}{2} \right) + \frac{(4n+1)\pi \, i}{2} $$ for positive integer $n.$ With this recipe, we get $$ n=9, \; \; \approx 4.062500618 + 58.11946410 \, i, $$ $$ n=10, \; \; \approx 4.165154772 + 64.40264941 \, i. $$

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Not as simple, but pretty good for 2-3 digits (and thus good enough for subsequent polishing with Newton-Raphson): $$\frac{((2k+1)\pi+i)\log((2k+1)\pi i)}{(2k+1)\pi}-(2k+1)\pi i$$ where $k$ is an integer. –  J. M. Jan 6 '12 at 2:39
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All values are $-W(-1)$, you just have to use all branches of the $W$ function.

$$\begin{align} & {}\ \vdots \\ -\mathrm{W}_{ -10 }(-1) & = 4.063742 + 58.049573 i \\ -\mathrm{W}_{ -9 }(-1) & = 3.949523 + 51.760122 i \\ -\mathrm{W}_{ -8 }(-1) & = 3.820554 + 45.469265 i \\ -\mathrm{W}_{ -7 }(-1) & = 3.672450 + 39.176440 i \\ -\mathrm{W}_{ -6 }(-1) & = 3.498515 + 32.880721 i \\ -\mathrm{W}_{ -5 }(-1) & = 3.287769 + 26.580472 i \\ -\mathrm{W}_{ -4 }(-1) & = 3.020240 + 20.272458 i \\ -\mathrm{W}_{ -3 }(-1) & = 2.653192 + 13.949208 i \\ -\mathrm{W}_{ -2 }(-1) & = 2.062278 + 7.588631 i \\ -\mathrm{W}_{ -1 }(-1) & = 0.318132 + 1.337236 i \\ -\mathrm{W}_{ 0 }(-1) & = 0.318132 -1.337236 i \\ -\mathrm{W}_{ 1 }(-1) & = 2.062278 -7.588631 i \\ -\mathrm{W}_{ 2 }(-1) & = 2.653192 -13.949208 i \\ -\mathrm{W}_{ 3 }(-1) & = 3.020240 -20.272458 i \\ -\mathrm{W}_{ 4 }(-1) & = 3.287769 -26.580472 i \\ -\mathrm{W}_{ 5 }(-1) & = 3.498515 -32.880721 i \\ -\mathrm{W}_{ 6 }(-1) & = 3.672450 -39.176440 i \\ -\mathrm{W}_{ 7 }(-1) & = 3.820554 -45.469265 i \\ -\mathrm{W}_{ 8 }(-1) & = 3.949523 -51.760122 i \\ -\mathrm{W}_{ 9 }(-1) & = 4.063742 -58.049573 i \\ -\mathrm{W}_{ 10 }(-1) & = 4.166242 -64.337984 i \\ & {}\ \vdots \end{align} $$

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