2022TrialShort answer

Q14 (b)

2 marks

Question

Simplify fully
$\frac1{1+\omega}+\frac1{1+\omega^2},$
given that $\omega$ is a non-real cube root of unity.

Try the question before revealing the answer.

Solution

$z^3=1$
$z^3-1=0$
$(z-1)(1+z+z^2)=0$

$1+\omega+\omega^2=0$

$\frac{1}{1+\omega}+\frac{1}{1+\omega^2}=\frac{1+\omega^2+1+\omega}{(1+\omega)(1+\omega^2)}$
$=\frac{1+\omega+\omega^2+1}{1+\omega+\omega^2+\omega^3}$
$=\frac{0+1}{0+\omega^3}$
Since $1+\omega+\omega^2=0$ ,

$=\frac{1}{\omega^3}=1$
Since $\omega^3=1$ .

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Syllabus alignment

Complex Numbers

Solving equations with complex numbersUse De Moivre's theorem and solve equations and polynomial problems involving complex and conjugate roots.

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