Complex solutions of $e^{e^z}=1$

Question

I know that the solution of $e^z = 1$ is $2\pi k$, for integers $k.$ Note $z$ is complex. I just don't really know how to start tackling $e^{e^z} = 1$

It seems that this means we want $e^z = 2\pi k$, i.e $z = \log(2\pi k)$, but this somehow seems like it's too simple to be right.

Answer 1

First off, there is an error in your question: if $\omega \in \mathbb{C}$, then $$ \mathrm{e}^{\omega} = 1 \implies \omega = 2\pi k\color{red}{i}, $$ where $k$ is any integer. Note that this is a purely imaginary number, rather than the real value $2\pi k$ in your question. Taking $\omega = \mathrm{e}^{z}$, we have $$ \mathrm{e}^{\mathrm{e}^z} = 1 \implies \mathrm{e}^z = 2\pi k i, $$ where $k$ is any integer.

Next, it might be helpful to (rather pedantically) expand out the right-hand side of the equation in terms of exponentials: $$ \mathrm{e}^z = 2\pi k i = (2\pi k) (i) = \mathrm{e}^{\log(2\pi k)} \mathrm{e}^{i\left(\frac{\pi}{2} + 2\pi\ell\right)} = \mathrm{e}^{\log(2\pi k) + i\left(\frac{\pi}{2} + 2\pi\ell\right)}, $$ where $\ell$ is an arbitrary integer and $\log$ is the the natural logarithm defined on $\mathbb{R}$; we are using Euler's formula to write out $i$ as an exponential expression. Now, equating the real and imaginary parts on the left- and right-hand sides of the equation, we have $$ \Re(z) = \log(2\pi k) \qquad\text{and}\qquad \Im(z) = \frac{\pi}{2} + 2\pi\ell, $$ where $k \in \mathbb{Z}\setminus\{0\}$ (note that the logarithm of zero is undefined, hence $k$ cannot be zero) and $\ell \in \mathbb{Z}$. In other words, with $k$ and $\ell$ as above, we conclude that $$ z = \log(2\pi k) + i \left( \frac{\pi}{2} + 2\pi\ell \right). $$

Answer 2

The solution should be for $k\neq 0$

$$e^z=2ik\pi\implies z=2ik_1\pi+\log 2ik\pi=2ik_1\pi+\log (2k\pi\cdot e^{\frac{\pi}2i+2ik\pi})=\\=2ik_1\pi+\log 2k\pi+\frac{\pi}2i+2ik\pi=\\=\log 2k\pi+\frac{\pi}2i+2in\pi$$

Refer here for Complex logarithm.

Answer 3

$e^{e^z} = 0$ can be rewritten as a system of two equations,

$$e^w = 1 \\ e^z = w$$

To solve the first equation, we rewrite $1+0i$ in the form $e^{\log(r) + i\theta} =re^{i\theta}$ using polar coordinates $(r,\theta)$. Obviously, $r = |1| = 1$ and so $\log(r)=0$. Also, $0/1 = \tan(\theta)$, so $\theta = (2\pi)k_1$, for any $k_1\in \mathbb{Z}$. Thus, $e^w = e^{2\pi k_1 i}$, so $w = 2\pi k_1 i$.

Solving the second equation is identical $-$ rewrite $0+2\pi k_1 i$ in the form $e^{\log(r)+i\theta}$. This gives us $r = |2\pi k_1 i| = 2\pi k_1$, so $\log(r) = \log(2\pi k_1)$. Then, $0/2\pi k_1 = \cot(\theta)$ so $\theta = \pi/2 + 2\pi k_2$ for any $k_2\in\mathbb{Z}$. Thus, the equation becomes $e^z = e^{\log(2\pi k_1) + i(\pi/2 + 2\pi k_2)}$, which gives the final answer,

$$ z = \log(2\pi k_1) + i(\pi/2 + 2\pi k_2) $$ for any $k_1,k_2 \in \mathbb{Z}$.

Answer 4

Given that we seek $z$ such that

$e^{e^z} = \exp(\exp(z)) = 1, \tag 1$

setting

$w = e^z = \exp(z), \tag 2$

we first must find $w$ satisfying

$\exp(w) = e^w = 1; \tag 3$

it is well-known that (3) binds precisely when

$w = 2 \pi i n, \; n \in \Bbb Z; \tag 4$

thus we seek $z$ obeying

$e^z = 2 \pi i n, \; n \in \Bbb Z; \tag 5$

we note, however, that for no $z \in \Bbb C$ do we have $e^z = 0$, though every other value is admissible; thus we refine (5) to read

$e^z = 2 \pi i n, \; n \in \Bbb Z \setminus \{0\}; \tag 6$

things get a little easier if we now express $z = x + iy$ in cartesian co-ordinates and $2 \pi i n$ in polars, thusly:

$2\pi i n = \vert 2\pi n \vert i = \vert 2\pi n \vert \exp(\pi i / 2), \; n > 0, \tag 7$

$2 \pi i n = -\vert 2 \pi n \vert i = \vert 2\pi n \vert \exp(3 \pi i / 2) \; n < 0; \tag 8$

then (6) breaks into the two equations

$e^z = e^{x + iy} = e^x e^{iy} = \vert 2\pi n \vert \exp(\pi i / 2), \; n > 0, \tag 9$

$e^z = e^{x + iy} = e^x e^{iy} = \vert 2\pi n \vert \exp(3 \pi i / 2), \; n < 0; \tag{10}$

(9) implies that

$e^x = \vert e^x e^{iy} \vert = \vert \vert 2\pi n \vert \exp(\pi i / 2) \vert = \vert 2 \pi n \vert, \tag{11}$

whence,

$n > 0: \; \; x = \ln \vert 2 \pi n \vert = \ln 2 \pi n, \; y = \dfrac{\pi i}{2} + 2\pi m, \; m \in \Bbb Z, \tag{12}$

and, via (10),

$n < 0: \; \; x = \ln \vert 2 \pi n \vert = \ln(-2 \pi n), \; y = \dfrac{3 \pi i}{2} + 2\pi m, \; m \in \Bbb Z; \tag{13}$

now with

$z = x + iy, \tag{14}$

(12) and (13) express the complete set of solutions to (1).

Note Added in Edit, Monday 9 April 2018 10:00 PM PST: One cannot help but wonder about extending this problem to, say, finding the solutions to

$\exp(\exp(\exp(z))) = 1, \tag{15}$

and so forth; indeed, defining

$E_1(z) = \exp(z), \tag{16}$

and, inductively,

$E_{n + 1}(z) = \exp(E_n(z)), \tag{17}$

so that

$E_2(z) = \exp(\exp(z)), \; E_3(z) = \exp(\exp(\exp(z))),\; E_4(z) = \exp(\exp(\exp(\exp(z)))), \tag{18}$

und so weiter, we are motivated to wonder just what the general pattern in solving

$E_n(z) = 1 \tag{19}$

might be? One thing appears certain, it will embrace a large number of integer parameters. But it appears finding the general formula will be a bit of a complex task, so I shall not attemp it here. Takers, anyone? End of Note.