2023TrialShort answer

Q12 (a) (i)

2 marks

Question

Let $\alpha$ be a real number and suppose that $z$ is a complex number such that

z+\frac{1}{z}=2\cos\alpha.

You may assume that

z^n+\frac{1}{z^n}=2\cos(n\alpha)

for all positive integer $n$ . Let

\omega=z+\frac{1}{z}.

Show that

\omega^4+\omega^3-3\omega^2-2\omega=z+\frac{1}{z}+z^2+\frac{1}{z^2}+z^3+\frac{1}{z^3}+z^4+\frac{1}{z^4}.

Try the question before revealing the answer.

Solution

(i) RTP:

\omega^4+\omega^3-3\omega^2-2\omega =z+\frac1z+z^2+\frac1{z^2}+z^3+\frac1{z^3}+z^4+\frac1{z^4}

\begin{aligned} \text{LHS}&=\left(z+\frac1z\right)^4+\left(z+\frac1z\right)^3-3\left(z+\frac1z\right)^2-2\left(z+\frac1z\right)\\ &=z^4+4z^2+6+\frac4{z^2}+\frac1{z^4}+z^3+3z+\frac3z+\frac1{z^3}-3\left(z^2+2+\frac1{z^2}\right)-2z-\frac2z\\ &=z^4+\frac1{z^4}+z^3+\frac1{z^3}+4z^2-3z^2+\frac4{z^2}-\frac3{z^2}+3z-2z+\frac3z-\frac2z+6-6\\ &=z^4+\frac1{z^4}+z^3+\frac1{z^3}+z^2+\frac1{z^2}+z+\frac1z\\ &=\text{RHS} \end{aligned}

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Complex Numbers

Arithmetic of complex numbersRepresent and calculate with complex numbers in Cartesian form, including conjugates, division, reciprocals and square roots.

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