Specialist Maths — Question 18

QCAA 2023, Paper 2 · 5 marks

Q18 · 2023 · Technology-activeComplex unfamiliar5 marks

QUESTION 18 (5 marks)

Consider the complex solutions to the following equation, where

0<\arg(z)<\pi

(z+1)(z^{14}-z^{13}+z^{12}-z^{11}+\cdots+z^4-z^3+z^2-z)=1-z.

Let

w_1

be the solution with the maximum possible real part and

w_2

be the solution with the maximum possible imaginary part. Show that

\frac{w_1^4}{w_2}\in\mathbb Z

WORKED SOLUTION

5 marks

ANSWER

\frac{w_1^4}{w_2}=1\in\mathbb Z

Worked solution

Simplifying the equation gives

z^{15}=1

. Therefore

z=\operatorname{cis}\left(\frac{2n\pi}{15}\right),

with

0<\arg z<\pi

. The solution with maximum real part is

w_1=\operatorname{cis}\left(\frac{2\pi}{15}\right),

and the solution with maximum imaginary part is

w_2=\operatorname{cis}\left(\frac{8\pi}{15}\right).

Hence

\frac{w_1^4}{w_2}=\frac{\operatorname{cis}(8\pi/15)}{\operatorname{cis}(8\pi/15)}=1\in\mathbb Z.

simplifies the original equation to

z^{15}=1

[1 mark]

determines the complex solutions subject to the argument restriction

[1 mark]

determines

w_1

[1 mark]

determines

w_2

[1 mark]

shows that

w_1^4/w_2

is an integer

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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