Show that $e^z = z + \lambda$ has exactly $m + n$ solution in a horizontal strip.

Asked Jan 31 '21 at 06:51

Active Feb 01 '21 at 01:33

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Suppose $\lambda \in \mathbb{C}$, show that for sufficient large $m$ and $n$, then the equation $e^z = z + \lambda$ has exactly $m + n$ solutions in the horizontal strip $\{- 2 \pi im < \operatorname{Im} z < 2 \pi i n\}$.

I tried to compute the increase in the argument of $f(z) = e^z - z + \lambda$ around the rectangle with vertices $R-2\pi im$, $R + 2 \pi in$, $-R + 2 \pi in$ and $-R + 2 \pi im$. When $R$ is sufficiently large, $f(x) \approx e^z$, but then I got that the increase in the argument around the rectangle is $4\pi (m + n)$. So what I did wrong?

edited Feb 01 '21 at 01:33

K.defaoite

12,536

asked Jan 31 '21 at 06:51

Andy

One needs to divide by $2\pi$ to convert from the increase in argument to the net number of zeros. (btw, it looks like your rectangle is oriented in the wrong direction) That leaves just a factor of $2$ to locate. – Greg Martin Jan 31 '21 at 07:04
@GregMartin Can you be more specific? According to my solution, there are 2(m+n) solutions in the strip. I don't know where is the extra 2 come from.... – Andy Jan 31 '21 at 07:07
When the real part of $z$ is $-R$ the exponential function $e^z$ is tiny. – Jyrki Lahtonen Jan 31 '21 at 07:20
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Does this answer your question? Showing that for large $m$ and $n$ the equation $e^z=z+\lambda$ has exactly $m+n$ solutions in the horizontal strip ${-2\pi*i m< Imz < 2i\pi n}$ – Jyrki Lahtonen Jan 31 '21 at 07:25
@JyrkiLahtonen I hope to use the argument principle to solve this problem, that link does not really solve my question. But your comment is very helpful though. – Andy Jan 31 '21 at 07:30

Show that $e^z = z + \lambda$ has exactly $m + n$ solution in a horizontal strip.

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