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Exercise 2.9 · Q25

Q.If ω=cis⁡2π3\omega=\operatorname{cis}\dfrac{2\pi}3, then the number of distinct roots of ∣z+1ωω2ωz+ω21ω21z+ω∣=0\begin{vmatrix}z+1&\omega&\omega^2\\\omega&z+\omega^2&1\\\omega^2&1&z+\omega\end{vmatrix}=0 is

(1) 11\n(2) 22\n(3) 33\n(4) 44
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We expand the 3×33\times3 determinant as a cubic polynomial in zz using ω=−12+32i\omega=-\dfrac12+\dfrac{\sqrt3}2i, simplify every coefficient with ω3=1\omega^3=1 and 1+ω+ω2=01+\omega+\omega^2=0, and then count the distinct roots of the resulting equation.

Step 1. Write the determinant with ω=−12+32i, ω2=−12−32i\omega=-\dfrac12+\dfrac{\sqrt3}2i,\ \omega^2=-\dfrac12-\dfrac{\sqrt3}2i.

D(z)=∣z+1ωω2ωz+ω21ω21z+ω∣.D(z)=\begin{vmatrix}z+1&\omega&\omega^2\\\omega&z+\omega^2&1\\\omega^2&1&z+\omega\end{vmatrix}.

Step 2. Expand along the first row.

D(z)=(z+1)[(z+ω2)(z+ω)−1]−ω[ω(z+ω)−ω2]+ω2[ω−ω2(z+ω2)].D(z)=(z+1)\big[(z+\omega^2)(z+\omega)-1\big]-\omega\big[\omega(z+\omega)-\omega^2\big]+\omega^2\big[\omega-\omega^2(z+\omega^2)\big].

Step 3. Expand (z+ω2)(z+ω)−1(z+\omega^2)(z+\omega)-1. Using ω⋅ω2=ω3=1\omega\cdot\omega^2=\omega^3=1 and ω+ω2=−1\omega+\omega^2=-1:

(z+ω2)(z+ω)−1=z2+z(ω+ω2)+ω3−1=z2−z+1−1=z2−z.(z+\omega^2)(z+\omega)-1=z^2+z(\omega+\omega^2)+\omega^3-1=z^2-z+1-1=z^2-z.

So the first bracket contributes (z+1)(z2−z)=z3−z2+z2−z=z3−z(z+1)(z^2-z)=z^3-z^2+z^2-z=z^3-z.

Step 4. Expand the second term. −ω[ω(z+ω)−ω2]=−ω[ωz+ω2−ω2]=−ω2z-\omega[\omega(z+\omega)-\omega^2]=-\omega[\omega z+\omega^2-\omega^2]=-\omega^2z. …

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