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IAL 2024 Jan Q2

A Level / Edexcel / FP2

IAL 2024 Jan Paper · Question 2

z=6−63i\begin{align*} z = 6 - 6\sqrt{3}\mathrm{i} \end{align*}

(a) (i) Determine the modulus of zz

(ii) Show that the argument of zz is −π3-\frac{\pi}{3}

(3)

Using de Moivre’s theorem, and making your method clear,

(b) determine, in simplest form, z4z^4

(2)

(c) Determine the values of ww such that w2=zw^2 = z , giving your answers in the form a+iba + ib , where aa and bb are real numbers.

(3)
解答

(a)(i)

Given z=6−63iz = 6 - 6\sqrt{3}\mathrm{i}, the modulus of zz is:

∣z∣= 62+(−63)2= 36+108= 144= 12\begin{align*} |z| = &\,\sqrt{6^2 + (-6\sqrt{3})^2}\\[4mm] = &\,\sqrt{36 + 108}\\[4mm] = &\,\sqrt{144}\\[4mm] = &\,12 \end{align*}

(a)(ii)

The argument of zz can be found using the inverse tangent:

arg⁡z= arctan⁡(−636)= arctan⁡(−3)= −π3\begin{align*} \arg z = &\,\arctan\left( \frac{-6\sqrt{3}}{6} \right)\\[4mm] = &\,\arctan(-\sqrt{3})\\[4mm] = &\,-\frac{\pi}{3} \end{align*}

(b)

Using De Moivre’s theorem, we first write zz in exponential or trigonometric form:

z= 12e−π3i\begin{align*} z = &\,12\mathrm{e}^{-\frac{\pi}{3}\mathrm{i}} \end{align*}

Then for z4z^4:

z4= (12e−π3i)4= 124e−4π3i= 20736(cos⁡(−4π3)+isin⁡(−4π3))= 20736(−12+i32)= −10368+103683i\begin{align*} z^4 = &\,\left( 12\mathrm{e}^{-\frac{\pi}{3}\mathrm{i}} \right)^4\\[4mm] = &\,12^4 \mathrm{e}^{-\frac{4\pi}{3}\mathrm{i}}\\[4mm] = &\,20736 \left( \cos\left(-\frac{4\pi}{3}\right) + \mathrm{i}\sin\left(-\frac{4\pi}{3}\right) \right)\\[4mm] = &\,20736 \left( -\frac{1}{2} + \mathrm{i}\frac{\sqrt{3}}{2} \right)\\[4mm] = &\,-10368 + 10368\sqrt{3}\mathrm{i} \end{align*}

(c)

We are given w2=z=12e−π3iw^2 = z = 12\mathrm{e}^{-\frac{\pi}{3}\mathrm{i}}. The values of ww are the square roots of zz. By De Moivre’s theorem:

w= (12ei(−π3+2kπ))12for k=0,1= 12ei(−π6+kπ)= ±23(cos⁡(−π6)+isin⁡(−π6))= ±23(32−12i)= ±(3−3i)\begin{align*} w = &\,\left( 12\mathrm{e}^{\mathrm{i}\left(-\frac{\pi}{3} + 2k\pi\right)} \right)^{\frac{1}{2}} \quad \text{for } k=0, 1\\[4mm] = &\,\sqrt{12} \mathrm{e}^{\mathrm{i}\left(-\frac{\pi}{6} + k\pi\right)}\\[4mm] = &\,\pm 2\sqrt{3} \left( \cos\left(-\frac{\pi}{6}\right) + \mathrm{i}\sin\left(-\frac{\pi}{6}\right) \right)\\[4mm] = &\,\pm 2\sqrt{3} \left( \frac{\sqrt{3}}{2} - \frac{1}{2}\mathrm{i} \right)\\[4mm] = &\,\pm (3 - \sqrt{3}\mathrm{i}) \end{align*}

So the values of ww are 3−3i3 - \sqrt{3}\mathrm{i} and −3+3i-3 + \sqrt{3}\mathrm{i}.