## IrishBoy123 one year ago complex

1. IrishBoy123

for |z| = 2, find M such that:- $\left| \frac{z^2 - 4z-3}{(z^2-7)(z^2+2} \right| \le M$

2. IrishBoy123

$\left| \frac{z^2 - 4z -3}{(z^2 - 7)(z^2 + 2)} \right|$ $= \frac{|z^2 - 4z -3|}{|z^2 - 7| \ |z^2 + 2|}$ from Triangle Inequality $| |z^2| - |(-7)| | \le |z^2 + (- 7)| \le |z^2| + |-7|$ $3 \le |z^2 - 7| \le 11$ $| |z^2| - |2| | \le |z^2 + 2| \le |z^2| + |2|$ $2 \le |z^2 - 2| \le 6$ $|z^2 - 4z - 3)| \le |z^2| + |-4z| + |-3| = 4 + 8 + 3 = 15$ $\frac{15}{3 \times 2} = \frac{5}{2} = M$