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.
Best Answer
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.