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Question: What is \[{\omega ^{100}} + {\omega ^{200}} + {\omega ^{300}}\] equal to, where \(\omega \) is the c...

What is ω100+ω200+ω300{\omega ^{100}} + {\omega ^{200}} + {\omega ^{300}} equal to, where ω\omega is the cube root of unity?
A) 11
B) 3ω3\omega
C) 3ω23{\omega ^2}
D) 00

Explanation

Solution

Hint : To solve this question, we will first use the condition that ω\omega is the cube root of unity. Using this we will make all possible equations which we can take so that we can solve the given expression. Now, after arranging the given expression into required form, we will get our answer out of four multiple choices given to us.

Complete step-by-step answer :
We have been given that ω\omega is the cube root of unity. We need to find the value of ω100+ω200+ω300{\omega ^{100}} + {\omega ^{200}} + {\omega ^{300}}.
So, it is given that, ω\omega is the cube root of unity.
i.e., ω=(1)13\omega = {(1)^{\dfrac{1}{3}}}
On taking cubes on both sides, we get
⇒(ω)3=((1)13)3 ⇒(ω)3=1......eq.(1) ⇒(ω)3−1=0 ⇒(ω)3−(1)3=0 ⇒(ω−1)[(ω)2+ω+1)]=0 ⇒(ω−1)=0,[(ω)2+ω+1)]=0.....eq.(2)  \Rightarrow {(\omega )^3} = {({(1)^{\dfrac{1}{3}}})^3} \\\ \Rightarrow {(\omega )^3} = 1......eq.(1) \\\ \Rightarrow {(\omega )^3} - 1 = 0 \\\ \Rightarrow {(\omega )^3} - {(1)^3} = 0 \\\ \Rightarrow (\omega - 1)[{(\omega )^2} + \omega + 1)] = 0 \\\ \Rightarrow (\omega - 1) = 0,[{(\omega )^2} + \omega + 1)] = 0.....eq.(2) \\\
Now, let us find the value of ω100+ω200+ω300{\omega ^{100}} + {\omega ^{200}} + {\omega ^{300}}. Firstly, we will modify the given expression ω100+ω200+ω300{\omega ^{100}} + {\omega ^{200}} + {\omega ^{300}} in the form of above solved equations.
=(ω3)33ω+(ω3)66ω2+(ω3)100= {({\omega ^3})^{33}}\omega + {({\omega ^3})^{66}}{\omega ^2} + {({\omega ^3})^{100}}
Now, using eq. (1),eq.{\text{ }}\left( 1 \right), we get
=(1)33ω+(1)66ω2+(1)100= {(1)^{33}}\omega + {(1)^{66}}{\omega ^2} + {(1)^{100}}
=(1)33ω+(1)66ω2+(1)100= {(1)^{33}}\omega + {(1)^{66}}{\omega ^2} + {(1)^{100}}
=ω+ω2+1= \omega + {\omega ^2} + 1
Now, using eq.(2),eq.\left( 2 \right), we get
=0= 0
So, ω100+ω200+ω300=0{\omega ^{100}} + {\omega ^{200}} + {\omega ^{300}} = 0

Note : Students might get confused with the statement, i.e., ω\omega is the cube root of unity. So, you can do one thing, just go with the words. i.e., by cube root of unity, they mean power 13\dfrac{1}{3} of 1.1. On the left side take omega, and on right side take cube root of unity. Now, we took a cube on both sides, so we got one cube of omega, and on the other side we got a cube of cube root of unity, which cancels the root, and we got 11 on the other side. Then, with that we solved further.