if $\ Re(z)<0$ and $e^z=z+1$ the $z=?$

Question

Is there a complex number $z$ such that $Re(z)<0$ and $e^z=z+1$?

Answer 1

Breaking up into real and imaginary components $z=x+iy$, we would have to prove that there exists $x,y$ such that

$$e^x\cos y = x+1$$

$$e^x\sin y = y$$

However, since $e^x < 1$, the second equation only has the solution $y=0$. But the real equation

$$e^x = x+1$$

has no solutions for $x<0$

Answer 2

Writing $z$ in it’s real and imaginary parts we have $z=a+bi$ with $a<0$ for a contradiction. Note that in particular $0\overset{*}{<}e^a\overset{**}{<}1.$ Now $$e^z=z+1 \implies e^{bi}=e^{-a}\left(z+1\right).$$ Now since $e^{bi}$ has a purely imaginary exponent, it’s modulus is $1$, for example via Euler’s formula. Thus $$1=||e^{bi}||\overset{*}{=}e^{-a}||z+1||\overset{**}{>}1\cdot ||z+1||,$$ which describes the locus of an (open) disc with centre $1$ and radius $1$. In particular, this contradicts the assumption the real part is negative since the entire locus has positive real part. This provides the contradiction we need to conclude there are no solutions $z\in \mathbb{C}$ with $Re(z)<0$ to $e^z=z+1$.