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I need to prove that for $\lambda\in \mathbb{C}$ and for $m,n\in\mathbb{Z}$ large enough, the equation: $$ e^z = z+\lambda$$ has exactly $m+n$ solutions $z$ such that $-2\pi m<\Im z<2\pi n$, where $\Im z$ denotes the imaginary part of $z$.

I thought about looking at the rectangle $\pm R + 2\pi in, \pm R - 2\pi im$ and try to use Rouche's Theorem with: $$f(z)=e^z-(z+\lambda)=P_{n+m}(z) + \sum_{k=m+n+1}^{\infty}\frac{z^k}{k!} - (z+\lambda)$$

where: $$P_{n+m}(z) =\sum_{k=0}^{m+n}\frac{z^k}{k!} $$

is a polynomial of degree $m+n$, and all is left is to prove that: $$|P_{n+m}(z)| > \left| \sum_{k=m+n+1}^{\infty}\frac{z^k}{k!} - (z+\lambda)\right|$$

in the boundary of the rectangle. But I don't know how to do that (or even if this solution is the right way to go)


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Try using the argument principle on your rectangle. – mrf Jan 9 '13 at 10:16
If you write $z = x + iy$ you obtain two equations, one for the modulus and the other for the argument of $e^z = e^x e^{iy}$. Then one variable can be eliminated and you obtain just one equation involving a tangent function from which the $2\pi$-periodicity is obvious. – Marek Jan 9 '13 at 10:27
The imaginary part $\Im z$ of a complex number $z$ is a real number. Modified the post accordingly. – Did Jan 9 '13 at 11:14

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