Dear friends,
I need help in showing the following.
$\displaystyle \lim_{\substack{\lambda\in\mathbb{R}\\ \lambda\to0}}\bigg(\frac{1}{\lambda}\log|1+z\lambd a|\bigg)=\mathrm{Re}(z)$
Thanks!
bkarpuz
Dear friends,
I need help in showing the following.
$\displaystyle \lim_{\substack{\lambda\in\mathbb{R}\\ \lambda\to0}}\bigg(\frac{1}{\lambda}\log|1+z\lambd a|\bigg)=\mathrm{Re}(z)$
Thanks!
bkarpuz
(you should specify that $\displaystyle \lambda\in\mathbb{R}$)
After expanding you get $\displaystyle |1+z\lambda|^2=1+2{\rm Re}(z)\lambda+o(\lambda)$ when $\displaystyle \lambda\to 0$, $\displaystyle \lambda\in\mathbb{R}$, from which the result follows quickly using $\displaystyle \log (1+u)=u+o(u)$ when $\displaystyle u\to 0$, and $\displaystyle \log u=\frac{1}{2}\log u^2$.
Actually, the limit can be computed when $\displaystyle \mathrm{Re}(\lambda)\neq0$.
In this case, we have $\displaystyle \lambda=r\mathrm{e}^{i\theta}$ with $\displaystyle \theta\neq\pm\pi/2$.
So that, for the function $\displaystyle f(z):=|z|^{-1}\log|1+z|$ for $\displaystyle z\in\mathbb{C}\backslash\{-1\}$, we have
$\displaystyle \lim_{r\to0^{+}}f(r\mathrm{e}^{i\theta})=\lim_{r\t o0^{+}}\frac{1}{2r}\log\big(1+2r\cos(\theta)+r^{2} \big)=\cos(\theta)$
by using the fact mentioned previously by Laurent ($\displaystyle \lim\nolimits_{\lambda\in\mathbb{R},\ \lambda\to0}f(\lambda)=1$).
On the other hand if $\displaystyle \mathrm{Re}(\lambda)=0$, it can be easily computed as
$\displaystyle \lim_{r\to0^{+}}f(r\mathrm{e}^{\pm i\pi/2})=\lim_{r\to0^{+}}\frac{1}{2r}\log\big(1+r^{2}\b ig)=0.$