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Use conjugacy only with real coefficients · source reference

[2017/IJC/Prelim/I/4] · Question pp. 3 · Working pp. 11, 12, 13

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The conjugate of a root need not be a root of a polynomial with complex coefficients.

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Question

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(a) w3+pw2+qw+30=0w^3+pw^2+qw+30=0, with p,qp,q real constants, has root w=2−iw=2-i. Find p,qp,q, showing working. [3]

(b) z2+(−5+2i)z+(21−i)=0z^2+(-5+2i)z+(21-i)=0 has root z=3+uiz=3+ui, where uu is real. Find uu and hence the second root in Cartesian form a+bia+bi, showing working. [5]

[2017/IJC/Prelim/I/4]

Source working

Printed working

(a) Method 1: Conjugate root 2+i2+i.

(w−(2−i))(w−(2+i))=(w−2+i)(w−2−i)=(w−2)2−i2=w2−4w+4+1=w2−4w+5.\begin{aligned}(w-(2-i))(w-(2+i))&=(w-2+i)(w-2-i)\\&=(w-2)^2-i^2=w^2-4w+4+1=w^2-4w+5.\end{aligned}

By inspection,

w3+pw2+qw+30=(w2−4w+5)(w+6).w^3+pw^2+qw+30=(w^2-4w+5)(w+6).

Coefficient of w2w^2: p=6−4=2p=6-4=2. Coefficient of ww: q=−24+5=−19q=-24+5=-19.

Method 2: Substitute w=2−iw=2-i (or 2+i2+i):

(2−i)3+p(2−i)2+q(2−i)+30=0,(2-i)^3+p(2-i)^2+q(2-i)+30=0, (3−4i)(2−i)+p(3−4i)+q(2−i)+30=0,(3-4i)(2-i)+p(3-4i)+q(2-i)+30=0, (6−3i−8i−4)+p(3−4i)+q(2−i)+30=0,(6-3i-8i-4)+p(3-4i)+q(2-i)+30=0, (32+3p+2q)+(−11−4p−q)i=0.(32+3p+2q)+(-11-4p-q)i=0.

Real: 3p+2q=−323p+2q=-32 (1). Imaginary: 4p+q=−114p+q=-11 (2). (1)−2(2)(1)-2(2): 3p−8p=−32+223p-8p=-32+22, so −5p=−10-5p=-10, p=2p=2. From (2), q=−11−4(2)=−19q=-11-4(2)=-19.

(b) Substitute z=3+uiz=3+ui:

(3+ui)2+(−5+2i)(3+ui)+(21−i)=0,(3+ui)^2+(-5+2i)(3+ui)+(21-i)=0, 9+6ui−u2−15−5ui+6i−2u+21−i=0,9+6ui-u^2-15-5ui+6i-2u+21-i=0, (15−2u−u2)+(u+5)i=0.(15-2u-u^2)+(u+5)i=0.

Imaginary part gives u+5=0⇒u=−5u+5=0\Rightarrow u=-5.

Margin alternative: Real part gives 15−2u−u2=015-2u-u^2=0, u2+2u−15=0u^2+2u-15=0, (u+5)(u−3)=0(u+5)(u-3)=0, hence u=−5u=-5 or 33. Reject 33 because u+5=3+5≠0u+5=3+5\ne0.

Method 1 for the second root: Let it be ww:

z2+(−5+2i)z+(21−i)=(z−3+5i)(z−w).z^2+(-5+2i)z+(21-i)=(z-3+5i)(z-w).

Compare the coefficient of zz: −5+2i=−w−3+5i⇒w=2+3i-5+2i=-w-3+5i\Rightarrow w=2+3i.

Method 2: Let the other root be a+bia+bi, a,b∈Ra,b\in\mathbb R.

(z−(3−5i))(z−(a+bi))=z2−(a+bi)z−(3−5i)z+(3−5i)(a+bi)=z2−[a+3+(b−5)i]z+(3−5i)(a+bi).\begin{aligned}(z-(3-5i))(z-(a+bi))&=z^2-(a+bi)z-(3-5i)z+(3-5i)(a+bi)\\&=z^2-[a+3+(b-5)i]z+(3-5i)(a+bi).\end{aligned}

Compare zz coefficients: −5+2i=−[a+3+(b−5)i]-5+2i=-[a+3+(b-5)i]. Real and imaginary parts give −(a+3)=−5⇒a=2-(a+3)=-5\Rightarrow a=2 and −(b−5)=2⇒b=3-(b-5)=2\Rightarrow b=3.

Teacher note

Because all coefficients in (a) are real, complex roots occur in conjugate pairs: 2+i2+i, conjugate of 2−i2-i, is a root. This theorem does not apply to the non-real coefficients in (b).

Printed answer key

(a) p=2,q=−19p=2,q=-19. (b) u=−5u=-5, second root 2+3i2+3i.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 3; solution pages 11, 12, 13. These are original document page numbers, retained as text metadata.