# Rouche's theorem application

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

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

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 $$0, hence $$f$$ has a zero for some $$z=-r$$ in the unit disc. By the above the zero is unique, so done!