# Constant Exponent

The equation in question is: ${ e }^{ x }=x$. It would be interesting to find a constant point in a function like exponent. Obviously there are no real solutions. But what if we consider complex numbers, are there any constant exponents?

We are looking for a complex number $z$ such as ${ e }^{ z }=z$. Let $z=x+iy$, where $x$ - real part, $y$ - imaginary part. Following the expanded definition of the exponent:

${ e }^{ x+iy }={ e }^{ x }\cdot (cos(y)+i\cdot sin(y))=x+iy$

Thus we get a system of two equations:

$\begin{cases} { x=e }^{ x }\cdot cos(y) \\ y={ e }^{ x }\cdot sin(y) \end{cases}$

Since there are no real solutions, $y\neq 0$ which also means $sin(y)\neq 0$. That means we can safely divide by $y$ and express $x$:

$\frac { x }{ y } =\frac { cos(y) }{ sin(y) }$

$x=y\cdot cot(y)$

Using this expression we get an equation that should help us get the value of $y$:

$y={ e }^{ y\cdot cot(y) }\cdot sin(y)$

Unfortunately, I wasn't able to derive a concrete root of this equation. However, let's consider these two functions:

$f(y)=\frac { y }{ sin(y) } \\ g(y)={ e }^{ y\cdot cot(y) }$

Both are even functions. If we build their graphs we get:

($f(y)$ - green, $g(y)$ - red)

As we can see there is an infinite number of intersections occurring roughly once in every $2\pi$ interval. That means there is an infinite number of fixed complex points who are equal to their exponent which I find fascinating.

If you have any idea how to derive $z$ or found an error, feel free to comment.

Note by Nick Kent
8 months, 1 week ago

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Interesting investigation. Starting with your system of two equations, I used multi-variate Newton Raphson iteration to find $x$ and $y$.
Here is the $z$ value with the smallest magnitude. Of course, there are other solutions as well:

$z \approx 0.3181 \pm 1.3372 i$

Python code is attached:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 import math import numpy as np import random S = 5.0 # Initial guesses x = -S + 2.0*S*random.random() y = -S + 2.0*S*random.random() # Iterative solution for j in range(0,100): f1 = math.exp(x) * math.cos(y) - x # functions from system f2 = math.exp(x) * math.sin(y) - y fvec = np.array([f1,f2]) J11 = math.exp(x) * math.cos(y) - 1.0 # Jacobian matrix entries J12 = -math.exp(x) * math.sin(y) J21 = math.exp(x) * math.sin(y) J22 = math.exp(x) * math.cos(y) - 1.0 J = np.array([[J11,J12],[J21,J22]]) vec = np.array([x,y]) right = np.dot(J,vec) - fvec Sol = np.linalg.solve(J, right) x = Sol[0] y = Sol[1] print x print y print "" print "" print "###############################" print "" 

- 8 months, 1 week ago

@Nick Kent Would you like to post this as a problem?

- 8 months, 1 week ago