Let ω be the complex number cos32π+isin32π. Then the number of distinct complex numbers z satisfying ∣∣z+1ωω2ωz+ω21ω21z+ω∣∣=0 is equal to.

ω=cos32π+isin32πApply C1→C1+C2+C3, we get∣∣zzzωz+ω21ω21z+ω∣∣=0⇒z∣∣111ωz+ω21ω21z+ω∣∣=0⇒z∣∣100ωz+ω2−ω1−ωω21−ω2z+ω−ω2∣∣=0by R2→R2−R1R3→R3−R1=(z+ω2−ω)(z+ω−ω2)−(1−ω)(1−ω2)=0⇒z2=0⇒ only one solution.

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Let $ω$ be the complex number $cos \frac{2 π}{3} + i sin \frac{2 π}{3}$ . Then the number of distinct complex numbers z satisfying $∣ ∣ z + 1 ω ω^{2} ω z + ω^{2} 1 ω^{2} 1 z + ω ∣ ∣ = 0$ is equal to.

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Charlotte Discussed

Let

ω

be the complex number

cos \frac{2 π}{3} + i sin \frac{2 π}{3}

. Then the number of distinct complex numbers z satisfying

∣ ∣ z + 1 ω ω^{2} ω z + ω^{2} 1 ω^{2} 1 z + ω ∣ ∣ = 0

is equal to.

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Text solutionVerified

ω = cos \frac{2 π}{3} + i sin \frac{2 π}{3}

Apply

C_{1} \to C_{1} + C_{2} + C_{3}

, we get

∣ ∣ z z z ω z + ω^{2} 1 ω^{2} 1 z + ω ∣ ∣ = 0

\Rightarrow z ∣ ∣ 111 ω z + ω^{2} 1 ω^{2} 1 z + ω ∣ ∣ = 0

\Rightarrow z ∣ ∣ 100 ω z + ω^{2} - ω 1 - ω ω^{2} 1 - ω^{2} z + ω - ω^{2} ∣ ∣ = 0

R_{2} \to R_{2} - R_{1}

R_{3} \to R_{3} - R_{1}

= (z + ω^{2} - ω) (z + ω - ω^{2}) - (1 - ω) (1 - ω^{2}) = 0

\Rightarrow z^{2} = 0

\Rightarrow

only one solution.

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Question Text	Let $ω$ be the complex number $cos \frac{2 π}{3} + i sin \frac{2 π}{3}$ . Then the number of distinct complex numbers z satisfying $∣ ∣ z + 1 ω ω^{2} ω z + ω^{2} 1 ω^{2} 1 z + ω ∣ ∣ = 0$ is equal to.
Answer Type	Text solution:1
Upvotes	150

Let ω be the complex number cos32π​+isin32π​. Then the number of distinct complex numbers z satisfying ∣∣​z+1ωω2​ωz+ω21​ω21z+ω​∣∣​=0 is equal to.

Text solutionVerified

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Let $ω$ be the complex number $cos \frac{2 π}{3} + i sin \frac{2 π}{3}$ . Then the number of distinct complex numbers z satisfying $∣ ∣ z + 1 ω ω^{2} ω z + ω^{2} 1 ω^{2} 1 z + ω ∣ ∣ = 0$ is equal to.