Find out the number of zeroes of $f(z) = e^z -3z -2 =0$ , inside $|z|=1.$

Question

In the first view I thought it to be a question of Rouche's theorem but when proceeded it was quiet tricky

As $f(z) = e^z- 3z- 2$

If we take $g(z)= - 3z-1$

Then $f(z) - g(z) = e^z - 2 = h(z)$ (say)

Now $|g(z)|= |-3z-1| > 3|z|-1 = 3-1 =2$ i.e $|g(z)|>2$ $(|z|=1)$

And $|h(z)|=|e^z-1|<e^{|z|}-1<e-1<2$

So $|h(z)|<2$ therefore

Since $|g(z)|>|h(z)|$

Therefore $Z(f)=Z(g)$

Which is given as

$g(z)=0$

i.e., $-3z-1=0 \implies z=-1/3$ is the required zero of the given function

And the no. Of zeroes of the fn $f(z)$ is $1$

Am I right kindly suggest

Answer 1

You have $$ f(z)\in -1-2z+z^2B(0,e^1-2), $$ $B(x,r)$ being the disk of radius $r$ around $x$.

This should be sufficient to apply the Rouché theorem, as $|1+2z|\ge 1>e-2$.