How many complex numbers \(z\) satisfy \[ |z|<2,\qquad \operatorname{Im}(z)>0,\qquad z^7+e^z=0? \]
Solution
Set \[ F(z)\coloneqq z^7+e^z. \]
1. The function \(F\) has exactly seven zeros in the disk\[ \abs{z}<2, \]counted with multiplicity.
On the circle \(\abs{z}=2\), \[ \abs{z^7}=2^7=128, \] while \[ \abs{e^z} = e^{\operatorname{Re}z} \leq e^2 < 128. \] Rouché’s theorem therefore shows that \(F(z)=z^7+e^z\) and \(z^7\) have the same number of zeros inside the circle. The latter has seven zeros counted with multiplicity, all at \(0\).
2. The function \(F\) has exactly one real zero, and it lies in\((-1,0)\).
For real \(x\), \[ F'(x)=7x^6+e^x>0, \] so \(F\) is strictly increasing on \(\RR\). Also \[ F(-1)=-1+e^{-1}<0 \] and \[ F(0)=1>0. \] The intermediate value theorem gives a real zero in \((-1,0)\), and strict monotonicity makes it unique.
3. Every zero of \(F\) is simple.
If \(z\) were a multiple zero, then \[ z^7+e^z=0 \] and \[ 7z^6+e^z=0. \] Thus \[ e^z=-z^7=-7z^6, \] so \[ z^6(z-7)=0. \] Hence \(z=0\) or \(z=7\). But \[ F(0)=1 \] and \[ F(7)=7^7+e^7\neq0. \] This contradiction proves simplicity.
4. Besides the one real zero from step 2, the disk\(\abs{z}<2\) contains six nonreal zeros, arranged in three conjugate pairs.
Steps 1 and 3 show that there are seven distinct zeros in the disk. Step 2 accounts for exactly one of them on the real axis. Since \[ \overline{F(z)} = F(\overline z), \] every nonreal zero occurs together with its complex conjugate. Thus the remaining six zeros form three conjugate pairs.
5. The number of solutions satisfying all three conditions is\[ \boxed{3}. \]
Each conjugate pair in step 4 contains exactly one zero with positive imaginary part and one with negative imaginary part. There are three such pairs.
6. Q.E.D.
Step 5 gives the requested number.