Complex number trig help

R

Run hard@thehsc

Well-Known Member

Dec 2, 2021

#1

How would you do 13 (a) and (b)?

D

$\therefore z^2(5z^2-11z^1 + 16 -11z^{-1} +5z^{-2})=0\\ \\ \therefore (5(z^2+z^{-2}) -11(z^1+z^{-1}) + 16 )= 0 $ if $ z \neq 0\\ \\ z=cis \theta \implies z^n = cis (n\theta )$ and $ z^{-n} = cis( {-n\theta} )\implies z^n+z^{-n} = 2cos(n\theta)\\ \\ \therefore 10cos(2\theta) -22cos(\theta )+ 16 = 0\\ \\ \therefore 5cos(\2\theta) -11cos \theta + 8 = 0 \\ \\ \therefore 5(2cos^2\theta-1)-11cos\theta+8 = 0\\ \\ \therefore 10cos^2\theta -11cos\theta+ 3 = 0 (\text {a quadratic eqn in } cos\theta) \\ \\ \implies cos\theta = \frac{1}{2}, \frac{3}{5} \implies sin\theta = \pm\frac{\sqrt 3}{2}, \pm \frac {4}{5} \\ \\ \therefore \text {The 4 roots are: }\frac{1}{2} \pm \frac{\sqrt 3}{2}i, \frac{3}{5} \pm \frac{4}{5}i$