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Worked solution

Factor a cubic and choose exact principal arguments

Do not use a calculator.

(a) Show that z=2iz=2i is a root of z3+2z+4i=0z^3+2z+4i=0. marks Hence find the other roots. marks

(b) Given w1=−1−iw_1=-1-i and w2=−62+22iw_2=-\dfrac{\sqrt6}{2}+\dfrac{\sqrt2}{2}i, find their moduli and arguments exactly. marks

[2021/RI/I/Q6(modified)]

CJC · Review problems · Question 17

Answer

(a) Other roots 1−i,−1−i1-i,-1-i. (b) ∣w1∣=∣w2∣=2|w_1|=|w_2|=\sqrt2; arguments −3π/4,5π/6-3\pi/4,5\pi/6.

(a) Verify and factor

(2i)3+2(2i)+4i=−8i+4i+4i=0(2i)^3+2(2i)+4i=-8i+4i+4i=0. Thus z−2iz-2i is a factor, and z3+2z+4i=(z−2i)(z2+2iz−2)z^3+2z+4i=(z-2i)(z^2+2iz-2). The quadratic roots are z=(−2i±−4+8)/2=−i±1z=(-2i\pm\sqrt{-4+8})/2=-i\pm1, so the other roots are 1−i1-i and −1−i-1-i.

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(b) Modulus and argument of w₁

∣w1∣=12+12=2|w_1|=\sqrt{1^2+1^2}=\sqrt2. Its point lies in quadrant III with reference angle π/4\pi/4. Using the principal-argument convention −π<arg⁡w≤π-\pi<\arg w\le\pi, arg⁡w1=−π+π/4=−3π/4\arg w_1=-\pi+\pi/4=-3\pi/4.

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(b) Modulus and argument of w₂

∣w2∣=6/4+2/4=2|w_2|=\sqrt{6/4+2/4}=\sqrt2. It lies in quadrant II, with reference angle tan⁡−1[(2/2)/(6/2)]=tan⁡−1(1/3)=π/6\tan^{-1}[(\sqrt2/2)/(\sqrt6/2)]=\tan^{-1}(1/\sqrt3)=\pi/6. Therefore arg⁡w2=π−π/6=5π/6\arg w_2=\pi-\pi/6=5\pi/6.

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Common mistakes

A reference angle alone is not the argument. The two points lie in different quadrants, and the principal argument of −1−i is negative.

Solution notes

Source filename says 2026; printed header says 2025/JC2. Both identities are retained; rendered pages determine radicals and modulus bars.

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