# Does this transcendental equation have solutions for a non real variable?

Having solved the transcendental equation $$e^{\frac{1}{\log(x)}}=x$$ I found that it has solutions for a real variable $$x.$$

Does it have solutions for not real $$x$$ (i.e. over the complexes, quaternions, octonions)?

Edit, June 24th 2020: I plotted $$f(x)=e^{\frac{1}{\log(x)}}-x$$ and saw many of these black dots which correspond to solutions for this online complex plotter (from my understanding). I don't know if the plotter is making a mistake or not...

I'm using an online complex plotter called: "The Complex Grapher" • Is that log base 10? May 2, 2019 at 4:16
• It should be log base e actually May 2, 2019 at 4:22

We can also show it analytically. Let $$x = re^{i\theta}$$ be a solution of the equation. Then, $$e^{\frac{1}{\ln(re^{i\theta})}} = re^{i\theta}$$ $$e^{1/(\ln(r)+i\theta)} = re^{i\theta}$$ $$\frac{1}{\ln(r)+i\theta} = \ln(r)+i\theta$$ $$\ln^2(r)+i2\ln(r)\theta -\theta^2 = 1$$ Equating imaginary parts implies that $$\theta = 0$$.

Hence, the solution must be real.

Domain coloring must be interpreted correctly. The best way to get a feel for how domain coloring works for a particular algorithm is to begin by plotting $$f(z) = z$$. By studying Figure 1 below we can see that the hue is given by the phase and the brightness by the modulus of $$f(z)$$. (We set the user-entered "magnitude modulus" parameter to the default value of 1/2.) Note that the brightness is independent of phase, that is, the brightness is circularly symmetric. The modulus of $$z$$ is unbounded. Brightness, however, varies between $$0$$ and $$1$$. There are many ways to deal with this. Let us focus on the dark bands. These correspond to values of $$f(z)$$ with an integer modulus. Note that in terms of finding the zeros of $$f(z)$$ that this convention is not very useful---we can't distinguish between locations where $$f(z)=0$$ and where $$|f(z)| \in \mathbb{Z}^+$$. For example, a region for which $$|f(z)|=7$$ is just as dark as a region for which $$f(z)=0$$. Thus, although $$f(z)=z$$ has only one zero, at $$z=0$$, the domain coloring plot indicates that there could be zeros for any $$z$$ for which $$|z|\in\mathbb{Z}$$. In terms of searching for zeros, the best way to interpret these dark regions is that they are regions where the function may have a zero.

In Figures 2 and 3 we give domain coloring plots of $$f(z) = e^{1/\log z}-z$$. Note that the zeros at $$z=e,1/e$$ are visibly indicated but that there are potentially many other zeros. It turns out that all the other dark regions correspond to regions where $$|f(z)| \approx n\in \mathbb{Z}^+$$. This can be most easily seen in a plot of $$|f(x+i y)|$$ as a function of $$x$$ and $$y$$. See Figure 5.

For large $$|z|$$ we have $$f(z) \approx 1-z.$$ In Figure 4 we give the domain coloring plot of this approximation. Note the similarity between Figures 3 and 4. This allows us to predict that the domain coloring plot of $$f(z)=e^{1/\log z}-z$$ will have dark bands for large $$|z|$$ when $$|1-z| = n\in\mathbb{Z}^+$$, that is, on circles with integer radius centered on $$(1,0)$$. This is clearly born out by Figure 3. Figure 1. Domain coloring plot of $$f(z)=z$$ with windowing $$[-4.945,4.945]\times[-3.330,3.330]$$. See https://talbrenev.com/complexgrapher/. Figure 2. Domain coloring plot of $$f(z) = e^{1/\log z}-z$$ with windowing $$[-4.945,4.945]\times[-3.330,3.330]$$. Figure 3. Domain coloring plot of $$f(z) = e^{1/\log z}-z$$ with windowing $$[-39.560,39.560]\times[-26.640,26.640]$$. Figure 4. Domain coloring plot of $$f(z) = 1-z$$ with windowing $$[-39.560,39.560]\times[-26.640,26.640]$$. Figure 5. Plot of the modulus of $$f(z)=e^{1/\log z}-z$$.

It does not seem to have complex solutions.

What I did was to consider the funtion $$F(a,b)=e^{\frac{1}{\log (a+i b)}}-(a+i b)$$ and define $$\Phi(a,b)=\Re(F(a,b))^2+\Im(F(a,b))^2$$ and made 3D and contour plots over quite large ranges $$(-100,+100)$$. The only apparent solutions correspond to $$b=0$$.