2023TrialShort answer

Q12 (a) (iii)

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}.

Hence or otherwise, find all solutions of

16(\cos\alpha)^4+8(\cos\alpha)^3-12(\cos\alpha)^2-4\cos\alpha+1=0.

Try the question before revealing the answer.

Solution

(iii)

16(\cos\alpha)^4+8(\cos\alpha)^3-12(\cos\alpha)^2-4\cos\alpha+1=0

(2\cos\alpha)^4+(2\cos\alpha)^3-3(2\cos\alpha)^2-2(2\cos\alpha)+1=0

\left(z+\frac1z\right)^4+\left(z+\frac1z\right)^3-3\left(z+\frac1z\right)^2-2\left(z+\frac1z\right)+1=0

\omega^4+\omega^3-3\omega^2-2\omega+1=0

z^4+\frac1{z^4}+z^3+\frac1{z^3}+z^2+\frac1{z^2}+z+\frac1z+1=0

z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1=0

For $z^9=1$ ,

z^9-1=0

(z-1)(z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1)=0

Therefore the solutions for $z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1=0$ are the complex solutions for $z^9=1$ .

\therefore\text{ solutions are: }\pm\frac{2\pi}{9},\ \pm\frac{4\pi}{9},\ \pm\frac{6\pi}{9},\ \pm\frac{8\pi}{9}

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Complex Numbers

Solving equations with complex numbersUse De Moivre's theorem and solve equations and polynomial problems involving complex and conjugate roots.
Other representations of complex numbersUse Euler's formula and convert between Cartesian, polar and exponential forms to solve problems.

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