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Complex Numbers question

2006 · Shift 0 · Q67
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Complex Numbers question

2006 · Shift 0 · Q67

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If z2+z+1=0{z^2} + z + 1 = 0z2+z+1=0, where z is complex number, then value of (z+1z)2+(z2+1z2)2+(z3+1z3)2+..........+(z6+1z6)2{\left( {z + {1 \over z}} \right)^2} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^2} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^2} + .......... + {\left( {{z^6} + {1 \over {{z^6}}}} \right)^2}(z+z1​)2+(z2+z21​)2+(z3+z31​)2+..........+(z6+z61​)2 is :
  1. A
    18
  2. B
    54
  3. C
    6
  4. D
    12
View written solutionFree

Correct answer: D

  1. Use the given equation

We are given z2+z+1=0.z^2+z+1=0.z2+z+1=0. Since z3−1=(z−1)(z2+z+1),z^3-1=(z-1)(z^2+z+1),z3−1=(z−1)(z2+z+1), and z2+z+1=0z^2+z+1=0z2+z+1=0, we get z3=1,z^3=1,z3=1, with z≠1z\neq 1z=1.

So zzz is a non-real cube root of unity: z=ω or ω2,z=\omega \text{ or } \omega^2,z=ω or ω2, where ω3=1,1+ω+ω2=0.\omega^3=1,\quad 1+\omega+\omega^2=0.ω3=1,1+ω+ω2=0.

Also, because z3=1z^3=1z3=1, 1z=z2.\frac{1}{z}=z^2.z1​=z2. More generally, 1zn=z3−nmod powers of z3=1,\frac{1}{z^n}=z^{3-n}\quad \text{mod powers of }z^3=1,zn1​=z3−nmod powers of z3=1, and in particular zn+1zn=zn+z−n.z^n+\frac{1}{z^n}=z^n+z^{-n}. zn+zn1​=zn+z−n.

  1. Find the repeating values of zn+1znz^n+\dfrac1{z^n}zn+zn1​

Let an=zn+1zn=zn+z−n.a_n=z^n+\frac1{z^n}=z^n+z^{-n}.an​=zn+zn1​=zn+z−n. Since z3=1z^3=1z3=1, values repeat with period 333.

For n=1n=1n=1

a1=z+1z=z+z2.a_1=z+\frac1z=z+z^2.a1​=z+z1​=z+z2. Using z2+z+1=0z^2+z+1=0z2+z+1=0, z+z2=−1.z+z^2=-1.z+z2=−1. So a1=−1  ⟹  a12=1.a_1=-1 \implies a_1^2=1.a1​=−1⟹a12​=1.

For n=2n=2n=2

a2=z2+1z2=z2+z.a_2=z^2+\frac1{z^2}=z^2+z.a2​=z2+z21​=z2+z. Again, z+z2=−1,z+z^2=-1,z+z2=−1, so a2=−1  ⟹  a22=1.a_2=-1 \implies a_2^2=1.a2​=−1⟹a22​=1.

For n=3n=3n=3

a3=z3+1z3=1+1=2,a_3=z^3+\frac1{z^3}=1+1=2,a3​=z3+z31​=1+1=2, thus a32=4.a_3^2=4.a32​=4.

Then the pattern repeats:

\quad a_5^2=a_2^2=1, \quad a_6^2=a_3^2=4.$$ --- 3. **Add all terms** We need $$a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+a_6^2.$$ Substitute the values: $$1+1+4+1+1+4=12.$$ --- 4. **Final answer** $$\boxed{12}$$ So the correct option is **D**. --- 5. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.
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