# Find the number of zeroes of $6z^3 + e^z + 1$ in the unit disc $|z|<1$

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.

• Umm, $e^0 = 1$, how did you get $|e^z| < 1$? On the unit disc you have $|e^z| < e$. Commented Jul 16, 2018 at 3:33
• Please use MathJax Commented Jul 16, 2018 at 3:36
• 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? Commented Jul 16, 2018 at 3:36