Roots of $e^z=1+z$ on complex plane

What are the roots in the complex plane of $e^z=1+z$?

Clearly $z=0$ is one root. On the real line, we can show that $e^x>1+x$ for all $x\neq 0$.

But what about the rest of the complex plane?

Solutions Collecting From Web of "Roots of $e^z=1+z$ on complex plane"

It is arguably hopeless to expect that the roots of the equation can be written down in closed form. There is still a lot we can say about them, though.

Some preliminary observations:

  • Since $\overline{e^z} = e^{\overline{z}}$ and $\overline{1+z} = 1 + \overline{z}$, it follows that if $z$ is a solution of
    e^z = 1 + z \tag{1}
    then so is $\overline{z}$.

  • Taking the absolute value of both sides of $(1)$ yields
    e^{\operatorname{Re}(z)} = |1+z|, \tag{2}
    so if $z$ is a root of $(1)$ and $|z| > 2$ then $e^{\operatorname{Re}(z)} > 1$ and thus $\operatorname{Re}(z) > 0$. In other words, the roots of $(1)$ with $|z| > 2$ lie in the right half-plane.

Now, by Hadamard’s factorization theorem, any entire function of order $1$ with finitely many zeros is of the form $e^{az}p(z)$, where $a \in \Bbb C$ and $p(z)$ is a polynomial. The function $e^z – 1 – z$ is entire and of order $1$ but is not of this form, so equation $(1)$ must have infinitely many roots.

The zeros of nonzero analytic functions may not accumulate, so in any bounded subset of $\Bbb C$ the equation may have only finitely many roots. This prompts us to investigate how the roots with large modulus behave.

Claim: if $\left(z_n = x_n + i y_n\right)$ is a sequence of roots of $(1)$ with $\lim_{n \to \infty} |z_n| = \infty$ then

\lim_{n \to \infty} \frac{x_n}{y_n} = 0.

Proof: Suppose to the contrary that we can find a $\delta > 0$ and a subsequence $\left(z_{n_k}\right)$ such that $x_{n_k} \geq \delta \left|y_{n_k}\right|$ for all $k$ (recall that $x_n > 0$ for $n$ large enough). Then

|z_{n_k}|^2 = x_{n_k}^2 + y_{n_k}^2 \leq \left(1 + \delta^{-2}\right) x_{n_k}^2,

so that $\lim_{k \to \infty} x_{n_k} = \infty$. Substituting $z_{n_k}$ into $(2)$ then yields

e^{x_{n_k}} &= \left|1 + z_{n_k}\right| \\
&\leq 1 + x_{n_k} + \left|y_{n_k}\right| \\
&\leq 1 + \left(1+\delta^{-1}\right) x_{n_k},

which is impossible for $k$ large enough since the left-hand side grows exponentially while the right-hand side grows linearly.


Consequently, if $(z_n = x_n + iy_n)$ is a sequence of roots of $(1)$ with $|z_n| \to \infty$ we have $z_n \sim iy_n$ as $n \to \infty$. So, for such a sequence, taking logarithms of both sides fo $(2)$ yields

x_n = \log|1 + z_n| \sim \log|z_n| \sim \log|y_n|

as $n \to \infty$. Therefore,

z_n \approx \log|y_n| + iy_n

as $n \to \infty$. We can interpret this to mean that, asymptotically, the large roots of $(1)$ lie on the line described parametrically by

z(t) = \log|t| + it,

where $t \in \Bbb R$ is large. In fact, if we assume $y>0$ and substitute $z \approx \log y + iy$ into $(1)$ we get

e^{\log y + iy} \approx 1 + z \approx z \approx iy

so that $e^{iy} \approx i$ and thus $y \approx 2\pi n + \tfrac{\pi}{2}$. This argument can be formalized, and he result is that

The large roots of the equation $e^z = 1 + z$ satisfy
z = \log\left(2\pi n + \pi/2\right) \pm i\left(2\pi n + \pi/2\right) + o(1)
as $n \to \infty$.

I would open it by denoting $z=a+bi$ and get:
$$e^z=1+z \\
e^{a+bi}=1+a+bi \\
e^a(\cos b+i \sin b)=1+a+bi $$
equlize real and imaginary parts gives :

  • $e^a\cos b=1+a,$
  • $e^a\sin b = b$

developing the first equation gives $b=\cos^{-1}\left(\frac{1+a}{e^a}\right)$. putting it in the second equation gives:
$$e^a\sin \cos^{-1}\left(\frac{1+a}{e^a}\right) = \cos^{-1}\left(\frac{1+a}{e^a}\right)$$
converting $\sin(\cos^{-1} x)=\sqrt{1-x^2}$ gives:
$$e^a\sqrt{1-\left(\frac{1+a}{e^a}\right)^2} = \cos^{-1}\left(\frac{1+a}{e^a}\right)$$
Hope it is helps.

EDIT:we can start with the second equation and get $a=\ln\left(\frac{b}{\sin b}\right)$. putting it in the first equation gives:
$$\frac{b}{\sin b}\cos b=1+\ln\left(\frac{b}{\sin b}\right)$$

Wolfram alpha gives a solution of $b = ±13.8790560027468…$ and $b = ±7.46148928565425…$. I can’t think about exact form of it.

Try Wolfram Alpha:

It involves the Lambert W function.