Question

Show that,

1. lim _z->infinity (𝑎𝑧+𝑏) ³/(𝑧+𝑑)³ 
2. lim _z->1 (1−z bar)/(1−z) does not exist

Answer 1

$1.\\ lim_{ z->\infin} \frac{(𝑎𝑧+𝑏)^3}{(𝑧+𝑑)^3 }\\ \text{apply l'hospital rule}\\ =lim_{ z->\infin} \frac{3a(𝑎𝑧+𝑏)^2}{3(𝑧+𝑑)^2}\\ =lim_{ z->\infin} \frac{a(𝑎𝑧+𝑏)^2}{(𝑧+𝑑)^2}\\ \text{apply l'hospital rule}\\ =lim_{ z->\infin} \frac{2a^2(𝑎𝑧+𝑏)}{2(𝑧+𝑑)}\\ =lim_{ z->\infin} \frac{a^2(𝑎𝑧+𝑏)}{(𝑧+𝑑)}\\ \text{apply l'hospital rule}\\ =lim_{ z->\infin} \frac{a^3}{1}\\ =a^3\\ 2.\\ lim_{ z->1} \frac{1-\bar z}{1-z}\\ =lim_{ (x,y)->(1,0)} \frac{1-x+iy}{1-x-iy}\\ \text{We choose a path such that it satisfy the point (1,0). Let x=my+1 be that path where m is a real number,}\\ =lim_{ y->0} \frac{1-my-1+iy}{1-my-1-iy}\\ =lim_{ y->0} \frac{(-m+i)y}{(-m-i)y}\\ =lim_{ y->0} \frac{(-m+i)}{(-m-i)}\\ =\text{doesn't exist}\\ \text{Since, for different value of m, we get different limits.}$