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I have studied Rouche's theorem and applied it to polynomial expressions but I don't seem to understand the problem in expressions with an exponential term. My approach to the above question is as follows-

Let $f(z) = 6z^3$ and $g(z) = e^z + 1$. Now $|e^z| < 1$ hence $|g(z)| < 2$ and $|f(z)| <6.$

Thus $|f(z)| > |g(z)|$ and so $f + g$ which is the required expression should have same number of zeros by Rouche's theorem which is $3$ as $6z^3$ has a zero at origin with multiplicity $3.$

I am afraid I might be fundamentally wrong somewhere. If yes, please explain where and how. Thanks.

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    $\begingroup$ Umm, $e^0 = 1$, how did you get $|e^z| < 1$? On the unit disc you have $|e^z| < e$. $\endgroup$
    – copper.hat
    Commented Jul 16, 2018 at 3:33
  • $\begingroup$ Please use MathJax $\endgroup$
    – saulspatz
    Commented Jul 16, 2018 at 3:36
  • $\begingroup$ Oh sorry. So |e^z| is less than or equal to 1? (I don't know how to write that in mathJax, any edit is welcome) but does that change the rest of my approach? $\endgroup$
    – deadcode
    Commented Jul 16, 2018 at 3:36
  • $\begingroup$ All I did to edit your code was to put '$' signs around what you had written. That's a big part of it. $\endgroup$
    – saulspatz
    Commented Jul 16, 2018 at 3:39
  • $\begingroup$ Well, your reasoning is not entirely clear. I will elaborate slightly in my answer. $\endgroup$
    – copper.hat
    Commented Jul 16, 2018 at 3:40

1 Answer 1

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Note that on the boundary of the disc you have $|f(z)|=|6 z^3| = 6$ and $|g(z)|=|e^z+1| \le e+1 < 4$.

Hence $|f(z)+g(z)-f(z)| = |g(z)| < 4 < 6 = |f(z)|$ on the unit disc (strict inequality is important) and so Rouché tells us that $f+g,f$ have the same number of zeros inside the disc.

Hence $f+g$ has 3 zeros inside the disc.

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  • $\begingroup$ So I was on the right track. Wolfram showed 1 root and hence I was not sure. Thanks for the help $\endgroup$
    – deadcode
    Commented Jul 16, 2018 at 3:48
  • $\begingroup$ And I had the idea $|e^z + 1|$ $\le$ $|e^z| + 1$ $\le$ $2$ . And 2<3 or 2 < 5 etc $\endgroup$
    – deadcode
    Commented Jul 16, 2018 at 3:50
  • $\begingroup$ @deadcode: The latter statement is incorrect as I pointed out in my first comment above. $\endgroup$
    – copper.hat
    Commented Jul 16, 2018 at 3:52
  • $\begingroup$ Wolfram just shows the real zero, there are two complex (conjugates) zeros. $\endgroup$
    – copper.hat
    Commented Jul 16, 2018 at 3:58
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    $\begingroup$ If you use Newtons method, you can approximate the zeros at $-0.63418$, $0.27726\pm i0.65950$. $\endgroup$
    – copper.hat
    Commented Jul 16, 2018 at 4:14

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