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below is a question I came across while studying for an exam, and I believe the way to handle it is to use the argument principle to count the increase in the argument, however, when I was reading an example of this in Gamelin's book, he only did an exampl which involved looking at arcs of circles and polynomials, which make this seems clear, but I've been unable to replicate his observations in my own setting. Would anyone mind giving me an outline of how you think one may approach this?

Question: How many zeroes does the function $f(z) = e^z + z$ have in the square $[-10,10] \times [-6 \pi i, 6 \pi i]$?

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  • $\begingroup$ It is not clear what you are asking. What observations are you referring to? $\endgroup$ – copper.hat Jul 14 '17 at 17:35
  • $\begingroup$ Ah, sorry. Gamelin does an example of a problem like this in his complex book. I was just referring to that. $\endgroup$ – mathgenesis22813 Jul 14 '17 at 17:36
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$f(z)$ is an entire function without double zeroes, since $f'(z)=e^z+1$.
The number of zeroes in the rectangle $R=[-10,10]\times [-6\pi i,6\pi i]$ is given by $$ N(R) = \frac{1}{2\pi i} \oint_{\partial R} \frac{f'(z)}{f(z)}\,dz = \frac{1}{2\pi i}\oint_{\partial R}\frac{1-z}{e^z+z}\,dz\tag{1}$$ where: $$ \int_{-10-6\pi i}^{10-6\pi i}\frac{1-z}{e^z+z}\,dz=\int_{-10}^{10}\frac{1-z+6\pi i}{e^z+z-6\pi i}\,dz\tag{B} $$ $$ \int_{10+6\pi i}^{-10+6\pi i}\frac{1-z}{e^z+z}\,dz=\int_{-10}^{10}\frac{-1+z+6\pi i}{e^z+z+6\pi i}\,dz\tag{T} $$ $$ \int_{10-6\pi i}^{10+6\pi i}\frac{1-z}{e^z+z}\,dz = \int_{-6\pi i}^{6\pi i}\frac{-9-z}{e^{z+10}+z+10}\,dz\tag{R} $$ $$ \int_{-10+6\pi i}^{-10-6\pi i}\frac{1-z}{e^{z}+z}\,dz = \int_{-6\pi i}^{6\pi i}\frac{z-11}{e^{z-10}+z-10}\,dz\tag{L} $$ where by numerical approximations we get that the contribute provided by the bottom and top side of $\partial R$ is $(B)+(T)\approx 4i$ and the contribute provided by the left and right side of $\partial R$ is $(L)+(R)\approx 40i$. It follows that $f$ has $\color{red}{7}$ zeroes inside $R$. These roots are given by the solutions of

$$\left\{\begin{array}{rcl}e^a \cos(b)+a &=& 0 \\ e^a\sin(b)+b &=& 0\end{array}\right. $$ with $a\in[-10,10]$ and $b\in[-6\pi,6\pi]$. From $\cos(b)=-ae^{-a}$ we get $a\geq -1$, so the original rectangle $R$ can be substantially resized, too. The problem boils down to counting the intersections between the blue curve and the purple curve in the range $(a,b)\in[-1,10]\times[-6\pi,6\pi]$:

$\hspace{3cm}$enter image description here

You can check I was not lying, they really are seven.

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  • $\begingroup$ How did you get the numertator in the integrand? I tried some attempts at substitution, but none have worked. $\endgroup$ – mathgenesis22813 Jul 14 '17 at 23:26
  • $\begingroup$ @m2271r: $$\oint_{\partial R}1\,dz = 0 $$ since $g(z)=1$ is an entire function. $\endgroup$ – Jack D'Aurizio Jul 14 '17 at 23:28
  • $\begingroup$ So $$\oint_{\partial R}\frac{e^z+1}{e^z+z}\,dz = \oint_{\partial R}\frac{1-z}{e^z+z}\,dz.$$ $\endgroup$ – Jack D'Aurizio Jul 14 '17 at 23:29
  • $\begingroup$ You can also exploit the fact that there is a real zero at $z=-W(1)\approx -0.56714329$. Here $W$ is Lambert's function. $\endgroup$ – Jack D'Aurizio Jul 14 '17 at 23:32

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