I must find the number of zeros of the function $$f(z)=\mathrm{log}(z+3)+z,$$ inside the disk $D_1(0)$, where $z \in \mathbb{C}$.

However I don't know how to apply Rouche's theorem: $|z|=1$ if $z\in\partial D_1(0)$, but $|\mathrm{log}(z+3)|$ is less than $1$ if $z=-1$ and is greater than $1$ if $z=1$ (clearly $-1,1\in\partial D_1(0))$. Can you give me a hint? Thank you in advance

  • $\begingroup$ Do you need to use Rouche? I can give you a fairly easy proof that there is a unique zero but with different methods $\endgroup$
    – Conrad
    Nov 25, 2019 at 3:06
  • $\begingroup$ Well, the exercise is in a section named "Applications of Rouche's theorem", but you can give me your proof anyway. Thank you $\endgroup$
    – Dr. Scotti
    Nov 25, 2019 at 7:06

1 Answer 1


With the caveat that the following doesn't use Rouche and I would welcome an easy solution using Rouche directly, a way to show that $f(z)=\mathrm{log}(z+3)+z$ has a unique (which turns out to be negative real) zero on the unit disk is:

Notice that $f'(z)=\frac{1}{z+3}+1$ and $|\frac{1}{z+3}| \le \frac{1}{2}$ on the unit disc, so $\Re f' \ge -\frac{1}{2}+1 >0$, hence $f$ is univalent on the unit disc (this is the well known Noshiro- Warshawski theorem and it is an easy exercise by integrating $f'$ on the segment connecting any two points in the unit disc), so $f$ has at most one zero in the unit disc.

But for $z=-r, 0 \le r \le 1$ obviously $g(r)=\log(3-r)-r$ satisfies, $g(0)=\log 3>0,g(1)=\log 2 - 1 <0$ so it has a zero for some $0<r<1$, hence $f$ has a zero for some $z=-r$ in the unit disc. By the above the zero is unique, so done!


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