I'm studying for my complex analysis qual, and I've found a problem that I haven't seen solved here yet:
Take $\lambda$ to be purely imaginary. Prove that $z-\lambda=\frac{1}{3}e^{z^2}$ has exactly one solution in the strip $S:=\{x+iy:x\in [-1,1]\}$ .
I've tried showing that the roots are all in the disk $|\lambda-z|=\frac{1}{3}|e^{z^2}|=\frac{1}{3}e^{x^2-y^2}\leq \frac {1}{3}e$, but after this, I have that on the circle $|\lambda-z|=\frac{1}{3}e$, $|\lambda-z|\geq \frac{1}{3}e^{z^2}$, where equality holds when the real part of $z$ is $\pm 1$ and the imaginary part is $0$. If I'm remembering correctly, the conditions for Rouché's Theorem require strict inequality.
How should I proceed?
On the circle $|\lambda-z|=\frac{1}{3}e\;$we have \begin{align*} & |\lambda-z| < 1 \\[4pt] \implies\;\;& |\text{Re}(\lambda-z)| < 1 \\[4pt] \implies\;\;& |\text{Re}(\lambda)-\text{Re}(z)| < 1 \\[4pt] \implies\;\;& |-\text{Re}(z)| < 1 \\[4pt] \implies\;\;& |\text{Re}(z)| < 1 \\[4pt] \end{align*} so we can't have $\text{Re}(z)=\pm 1$.