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?
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
$$
\begin{align}
e^{x_{n_k}} &= \left|1 + z_{n_k}\right| \\
&\leq 1 + x_{n_k} + \left|y_{n_k}\right| \\
&\leq 1 + \left(1+\delta^{-1}\right) x_{n_k},
\end{align}
$$
which is impossible for $k$ large enough since the left-hand side grows exponentially while the right-hand side grows linearly.
Q.E.D.
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 :
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:
https://www.wolframalpha.com/input/?i=solve+e%5Ex+%3D+1%2Bx
It involves the Lambert W function.