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In an exercise I am required to show that the only solution to $e^z+z-a=0$ in the right half plane is a real one, where $a$ is fixed a real number larger than 1. I am not able to work it out. I tried to used the Argument Principle to compute the number of zeros of this function, but I cannot make my integral converge. Could anybody give me a hand?

Thanks very much!

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    This is a standard problem in applications of Rouche but it is usually given as $e^{-z} +z -a =0$.2012-07-24

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Let $a=\log(3\pi/2)$, so $a$ is real and $a\gt1$. Let $z=\log(3\pi/2)+(3\pi/2)i$, so the real part of $z$ is positive, and $z$ is not real. Then $\eqalign{e^z+z&=e^{\log(3\pi/2)}e^{(3\pi/2)i}+z=(3\pi/2)(\cos(3\pi/2)+i\sin(3\pi/2))+z\cr&=-(3\pi/2)i+\log(3\pi/2)+(3\pi/2)i=a\cr}$

So, for $a=\log(3\pi/2)$, there is a nonreal solution in the right half-plane.

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    @GerryMyerson o.k. I got it. See my comment above for the correct formulation of the question.2012-07-26
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I tried $a=2$. Solutions of $e^z+z-2=0$ are $-W_k(e^2)+2$, where $W_k$ are the branches of the Lambert $W$ function. The whole doubly-infinite sequence of solutions is in the right half-plane.

Indeed, is there any value of $a>1$ where there is a unique solution in the right half-plane?