Complex Numbers And Quadratic Equations question 674
Question: If $ z^{2}+z+1=0, $ where z is complex number, then the value of $ {{( z+\frac{1}{z} )}^{2}}+{{( z^{2}+\frac{1}{z^{2}} )}^{2}}+{{( z^{3}+\frac{1}{z^{3}} )}^{2}} $ $ +….+{{( z^{6}+\frac{1}{z^{6}} )}^{2}} $ is
Options:
A) 18
B) 54
C) 6
D) 12
Show Answer
Answer:
Correct Answer: D
Solution:
$ {z^{2}} + z +1 = 0 \Rightarrow z = \omega or {{\omega }^{2}} $ So, $ z+\frac{1}{z}=\omega +{{\omega }^{2}} =-1 $ $ {z^{2}}+\frac{1}{z^{2}}={{\omega }^{2}}+\omega =-1 $ $ {z^{3}}+\frac{1}{z^{3}}={{\omega }^{3}}+{{\omega }^{3}} =2 $ $ {z^{4}}+\frac{1}{z^{4}}=-1,z^{5}+\frac{1}{z^{2}}=-1andz^{6}+\frac{1}{z^{6}}=2 $
$ \therefore ~The given sum =1+1+4+1+1+4=12 $
BETA
