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Worked solution

Use conjugate roots and a reciprocal substitution

The complex numbers z1=2z_1=2 and z2=iz_2=i are roots of z3+Az2+Bz+C=0z^3+Az^2+Bz+C=0, where A,B,CA,B,C are real.

(i) State the third root and find A,B,CA,B,C. marks

(ii) Hence, with a suitable substitution, find the roots of 2w3−w2+2w−1=02w^3-w^2+2w-1=0. marks

[2013/SRJC/I/4]

CJC · Review problems · Question 4

Answer

(i) Third root −i-i; A=−2A=-2, B=1B=1, C=−2C=-2. (ii) w=12,i,−iw=\tfrac12,i,-i.

(i) Use the conjugate-root property

The coefficients are real, so the conjugate −i-i is the third root. The monic cubic is (z−2)(z−i)(z+i)=(z−2)(z2+1)=z3−2z2+z−2(z-2)(z-i)(z+i)=(z-2)(z^2+1)=z^3-2z^2+z-2. Therefore A=−2A=-2, B=1B=1, C=−2C=-2.

How did this part go?

(ii) Convert reciprocal roots

The new equation has no zero root because its constant term is −1-1. Put w=1/zw=1/z with z≠0z\ne0. Then 2/z3−1/z2+2/z−1=02/z^3-1/z^2+2/z-1=0. Multiplying by −z3-z^3 gives z3−2z2+z−2=0z^3-2z^2+z-2=0, whose roots are 2,i,−i2,i,-i. Their reciprocals are 1/2,−i,i1/2,-i,i, so w=1/2,i,−iw=1/2,i,-i. The source writes the same reciprocal transformation by first rearranging the fractions.

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Common mistakes

Real coefficients guarantee the conjugate root; this need not hold for complex coefficients. Exclude zero before using a reciprocal substitution.

Solution notes

Source filename says 2026, but its printed header says 2025/JC2; both are retained without silently changing the source identity.

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