De Moivre’s Theorem (LAQs)

Maths-2A | 2. De Moivre’s Theorem – LAQs:
Welcome to LAQs in Chapter 2: De Moivre’s Theorem. This page contains the most Important FAQs for Long Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.

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LAQ-1 : If n is a positive integer then show that (1+i)²ⁿ + (1-i)²ⁿ = 2⁽ⁿ⁺¹⁾ cos⁡(nπ/2)

First we find the mod-amp form of $$1+i$$

Let $$x + iy = 1 + i$$ $$x = 1$$ $$y = 1$$

$$r = \sqrt{x^2 + y^2} = \sqrt{1^2 + 1^2} =\sqrt{2}$$

$$\tan\theta = \frac{y}{x} = 1/1 = \tan \frac{\pi}{4}$$

$$\theta = \frac{\pi}{4}$$

mod-Amp form $$1+i$$

$$r(\cos\theta + i \sin\theta) = \sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$$

$$(1 + i)^{2n} = (\sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}))^{2n}$$

$$= (\sqrt{2})^{2n} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{2n}$$

$$= 2^n (\cos(2n) \frac{\pi}{4} + i \sin(2n) \frac{\pi}{4})$$

$$= 2^n (\cos n\pi/2 + i \sin n\pi/2) … (1)$$

Similarly,

$$(1-i)^{2n} = 2^n(\cos n\pi/2 – i\sin n\pi/2) … (2)$$

Adding (1) & (2), we get

$$(1+i)^{2n} + (1-i)^{2n}$$

$$= 2^n ((\cos n\pi/2 + i\sin n\pi/2) + (\cos n\pi/2 – i\sin n\pi/2))$$

$$= 2^n (2\cos n\pi/2) = 2^{n+1}\cos n\pi/2$$

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LAQ-2 : If n is a positive integer then show that (1+i)ⁿ + (1-i)ⁿ = 2^(n+2)/2) cos⁡(nπ/4)

First we find the mod-amp form of $$1+i$$

Let $$x + iy = 1 + i$$ $$x = 1$$ $$y = 1$$

$$r = \sqrt{x^2 + y^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$$

$$\tan \theta = \frac{y}{x} = 1/1 = 1 = \tan \frac{\pi}{4}$$

$$\theta = \frac{\pi}{4}$$

mod-Amp form of $$1+i$$

$$r(\cos \theta + i \sin \theta) = \sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$$

$$(1+i)^n = (\sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}))^n$$

$$= (\sqrt{2})^n (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^n$$

$$= 2^{n/2} (\cos(n) \frac{\pi}{4} + i\sin(n) \frac{\pi}{4})$$

$$= 2^{n/2} (\cos n\pi/4 + i\sin n\pi/4) \ … (1)$$

Similarly,

$$(1-i)^n = 2^{n/2} (\cos n \pi/4 – i \sin n \pi/4) \ … (2)$$

Adding (1) & (2), we get

$$(1+i)^n + (1-i)^n$$

$$= 2^{n/2} ((\cos n \pi/4 + i \sin n \pi/4) + (\cos n \pi/4 – i\sin n \pi/4))$$

$$= 2^{n/2} (2\cos n\pi/4) = 2^{n/2+1}\cos n\pi/4 = 2^{n+2/2}\cos n\pi/4$$

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LAQ-3 : If n is a positive integer then show that (1+cosθ+isinθ)ⁿ +(1+cosθ-isinθ)ⁿ = 2ⁿ⁺¹ cosⁿ (θ/2) cos⁡(nθ/2)

First we find the mod-amp form of $$1 + \cos\theta + i\sin\theta$$

$$1 + \cos\theta + i\sin \theta = 2 \cos^2 \frac{\theta}{2} + i2\sin \frac{\theta}{2} \cos \frac{\theta}{2} = 2 \cos\frac{\theta}{2}(\cos \frac{\theta}{2} + i\sin \frac{\theta}{2})$$

$$(1 + \cos\theta + i \sin\theta)^n = (2 \cos \frac{\theta}{2}(\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}))^n = 2^n \cos^n \frac{\theta}{2}(\cos n \frac{\theta}{2} + i \sin n \frac{\theta}{2}) … (1)$$

Similarly, $$(1 + \cos \theta – i \sin \theta)^n = 2^n \cos^n \frac{\theta}{2}(\cos n \frac{\theta}{2} – i \sin n \frac{\theta}{2}) … (2)$$

Adding (1) & (2), $$(1 + \cos \theta + i \sin \theta)^n + (1 + \cos \theta – i \sin \theta)^n$$

$$= 2^n \cos^n \frac{\theta}{2} ((\cos n \frac{\theta}{2} + i \sin n \frac{\theta}{2}) + (\cos n \frac{\theta}{2} – i \sin n \frac{\theta}{2}))$$

$$= 2^n \cos^n \frac{\theta}{2} (2 \cos n\frac{\theta}{2}) = 2^{n+1} \cos^n \frac{\theta}{2}\cos n\frac{\theta}{2}$$

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LAQ-4 : If m,n are integers and x = cosα + i sinα, y = cosβ + i sinβ, then prove that x^m yⁿ + 1/(x^m yⁿ) = 2cos⁡(mα+nβ) and x^m yⁿ – 1/(x^m yⁿ) = 2isin(mα+nβ)

Given $$x = \cos \alpha + i \sin \alpha$$ $$y = \cos \beta + i \sin \beta$$

$$x^m = (\cos \alpha + i \sin \alpha)^m = \cos m\alpha + i \sin m\alpha = \text{cis} m\alpha$$

$$y^n = (\cos \beta + i \sin \beta)^n = \cos n\beta + i \sin n\beta = \text{cis} n\beta$$

$$x^my^n = (\text{cis} m\alpha)(\text{cis} n\beta) = \text{cis}(m\alpha + n\beta) = \cos(m\alpha+n\beta) + i \sin(m\alpha + n\beta) \ … \ (1)$$

Hence $$1/x^my^n = \cos (m\alpha+n\beta) – i \sin (m\alpha+n\beta) \ … \ (2)$$

By adding (1) and (2), we get $$x^my^n + 1/x^my^n = 2\cos(m\alpha+n\beta)$$

By subtracting (2) from (1), we get $$x^my^n – 1/x^my^n = 2i \sin(m\alpha+n\beta)$$

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LAQ-5 : If α,β are roots of the equation x² – 2x + 4 = 0, then show that αⁿ + βⁿ = 2ⁿ⁺¹ cos(nπ/3)

Given $$x^2 – 2x + 4 = 0$$

$$x = 2 \pm \sqrt{4 – 4(1)(4)}/2(1) = 2 \pm \sqrt{-12}/2 = 1 \pm i\sqrt{3}$$

Now, we find the mod-amp form of $$1 + i\sqrt{3}$$

Let $$x + iy = 1 + i\sqrt{3}$$ $$x = 1$$ $$y = \sqrt{3}$$

$$r = \sqrt{x^2 + y^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$$

Also, $$\tan\theta = y/x = \sqrt{3}/1 = \tan \pi/3$$

$$\theta = \pi/3$$

mod-Amp form of $$1 + i\sqrt{3}$$ is $$r(\cos\theta + i\sin \theta) = 2(\cos \pi/3 + i\sin \pi/3)$$

$$(1 + i\sqrt{3})^n = (2(\cos \pi/3 + i\sin \pi/3))^n$$

$$= (2)^n(\cos \pi/3 + i\sin \pi/3)^n = 2^n(\cos n \pi/3 + i \sin n \pi/3) \ … \ (1)$$

Similarly,

$$(1 – i\sqrt{3})^n = 2^n (\cos n \pi/3 – i \sin n \pi/3) \ … \ (2)$$

Adding (1) & (2), we get $$\alpha^n + \beta^n = (1 + i\sqrt{3})^n + (1 – i\sqrt{3})^n$$

$$= 2^n ((\cos n \pi/3 + i \sin n \pi/3) + (\cos n \pi/3 – i \sin n \pi/3)) = 2^n (2\cos n\pi/3) = 2^{n+1}\cos n\pi/3$$

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LAQ-6 : If n is a positive integer then S.T (p+iq)^1/n + (p-iq^1/n = 2(p² + q²)^1/2n cos(1/n Tan^-1 q/p)

First we reduce $$p + iq$$ $$x = p$$ $$y = q$$

$$r = \sqrt{x^2 + y^2} = \sqrt{p^2 + q^2}$$

$$\tan \theta = y/x$$ $$\theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{q}{p}$$

Mod-amp form of $$p + iq$$ $$r(\cos\theta + i\sin \theta) = \sqrt{p^2 + q^2} (\cos(\tan^{-1} \frac{q}{p}) + i\sin(\tan^{-1} \frac{q}{p}))$$

$$(p + iq)^{1/n} = (\sqrt{p^2 + q^2} (\cos(\tan^{-1} \frac{q}{p}) + i\sin(\tan^{-1} \frac{q}{p})))^{1/n}$$

$$= (p^2 + q^2)^{1/2n} (\cos \frac{1}{n} (\tan^{-1} \frac{q}{p}) + i\sin \frac{1}{n} (\tan^{-1} \frac{q}{p})) \ … \ (1)$$

Similarly, $$(p-iq)^{1/n} = (p^2 + q^2)^{1/2n} (\cos \frac{1}{n} (\tan^{-1} \frac{q}{p}) – i\sin \frac{1}{n} (\tan^{-1} \frac{q}{p})) \ … \ (2)$$

Adding (1) & (2), $$(p + iq)^{1/n} + (p – iq)^{1/n} = (p^2 + q^2)^{1/2n} (2\cos \frac{1}{n} \tan^{-1} \frac{q}{p})$$

$$= 2(p^2 + q^2)^{1/2n} \cos(\frac{1}{n} \tan^{-1} \frac{q}{p})$$

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LAQ-7 : Show that one value of (1+sin π/8+icos π/8)/(1+sin π/8-icos π/8)^8/3 = -1

Let $$z = \sin \frac{\pi}{8} + i \cos \frac{\pi}{8}$$

Then $$1/z = \frac{1}{\sin \frac{\pi}{8} + i \cos \frac{\pi}{8}} = \frac{1(\sin \frac{\pi}{8} – i \cos \frac{\pi}{8})}{(\sin \frac{\pi}{8} + i \cos \frac{\pi}{8})(\sin \frac{\pi}{8} – i \cos \frac{\pi}{8})} = \frac{\sin \frac{\pi}{8} – i \cos \frac{\pi}{8}}{\sin^2 \frac{\pi}{8} + \cos^2 \frac{\pi}{8}} = (\sin \frac{\pi}{8} – i \cos \frac{\pi}{8})$$

$$\left(1 + \frac{\sin \frac{\pi}{8} + i \cos \frac{\pi}{8}}{1} + \frac{\sin \frac{\pi}{8} – i \cos \frac{\pi}{8}}{1}\right)^{8/3} = \left(1 + z + \frac{1}{z}\right)^{8/3} = z^{8/3}$$

$$= (\sin \frac{\pi}{8} + i \cos \frac{\pi}{8})^{8/3} = [\cos(\frac{\pi}{2} – \frac{\pi}{8}) + i \sin (\frac{\pi}{2} – \frac{\pi}{8})]^{8/3}$$

$$= (\cos 4\pi – \frac{\pi}{8} + i \sin 4\pi – \frac{\pi}{8})^{8/3} = (\cos \frac{3\pi}{8} + i \sin \frac{3\pi}{8})^{8/3} = (\cos \frac{8}{3}(3\pi/8) + i \sin \frac{8}{3} (3\pi/8))$$

$$= \cos\pi + i\sin\pi = \cos180^\circ + i\sin180^\circ = -1 + i(0) = -1$$

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LAQ-8 : If n is an integer and z=cisθ then show that (z²ⁿ-1)/(z²ⁿ+1)=i tan ⁡nθ

Given $$z = \text{cis}\ \theta = \cos \theta + i\sin \theta$$

$$L.H.S = z^{2n} – \frac{1}{z^{2n}} + 1 = (\cos\theta + i \sin\theta)^{2n} – \frac{1}{(\cos\theta + i \sin\theta)^{2n}} + 1$$

$$= \cos(2n)\theta + i\sin(2n)\theta – \frac{1}{\cos(2n)\theta + i\sin(2n)\theta} + 1$$

$$= -\left(1-\cos 2n\theta\right) + i \sin 2n\theta / \left(1 + \cos 2n\theta\right) + i \sin 2n\theta$$

$$= -\left(2\sin^2 n\theta\right) + i(2\sin n\theta \cos n\theta) / 2\cos^2 n\theta + i(2\sin n\theta \cos n\theta)$$

$$= 2 i^2 \sin^2 n\theta + i(2\sin n\theta \cos n\theta) / 2\cos^2 n\theta + i(2\sin n\theta \cos n\theta)$$

$$= 2 i \sin n\theta (i \sin n\theta + \cos n\theta) / 2 \cos n\theta (\cos n\theta + i \sin n\theta) = i \tan n\theta = R.H.S$$

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LAQ-9 : If cosα+cosβ+cosγ=0=sinα+sinβ+sinγ, then show that cos²α+cos²β+cos²γ=3/2=sin²α+sin²β+sin²γ

Let $$a = \cos\alpha + i\sin\alpha = \text{cis}\alpha$$ $$b = \cos\beta + i\sin\beta = \text{cis}\beta$$ $$c = \cos\gamma + i\sin\gamma = \text{cis}\gamma$$

Then we have:

$$a + b + c = (\cos\alpha + i\sin\alpha) + (\cos\beta + i\sin\beta) + (\cos\gamma + i\sin\gamma)$$

$$= (\cos\alpha + \cos\beta + \cos\gamma) + i(\sin\alpha + sin\beta + sin\gamma)$$

$$= 0 + i(0) = 0$$

$$a + b + c = 0$$

$$(a + b + c)^2 = 0$$

$$\Rightarrow a^2 + b^2 + c^2 + 2(ab + bc + ca) = 0$$

$$\Rightarrow a^2 + b^2 + c^2 + 2abc\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = 0$$

$$\Rightarrow a^2 + b^2 + c^2 + 2abc\left(\frac{1}{\cos\alpha} + \frac{1}{\text{cis}\beta} + \frac{1}{\text{cis}\gamma}\right) = 0$$

$$\Rightarrow a^2 + b^2 + c^2 + 2abc(\text{cis}(-\alpha) + \text{cis}(-\beta) + \text{cis}(-\gamma)) = 0$$

$$\Rightarrow a^2 + b^2 + c^2 + 2abc[(\cos\alpha + \cos\beta + \cos\gamma) – i(\sin\alpha + sin\beta + sin\gamma)] = 0$$

$$\Rightarrow a^2 + b^2 + c^2 + 2abc(0 – i(0)) = 0$$

$$\Rightarrow a^2 + b^2 + c^2 + 2abc(0) = 0$$

$$\Rightarrow a^2 + b^2 + c^2 = 0$$

$$\Rightarrow (\text{cis}\alpha)^2 + (\text{cis}\beta)^2 + (\text{cis}\gamma)^2 = 0$$

$$\Rightarrow \text{cis}2\alpha + \text{cis}2\beta + \text{cis}2\gamma = 0$$

$$\Rightarrow (\cos 2\alpha + i\sin 2\alpha) + (\cos 2\beta + i\sin 2\beta) + (\cos 2\gamma + i\sin 2\gamma) = 0$$

$$\Rightarrow (\cos 2\alpha + cos 2\beta + cos 2\gamma) + i(sin 2\alpha + sin 2\beta + sin 2\gamma) = 0 + i(0)$$

Equating the real parts, we get $$\cos 2\alpha + \cos 2\beta + \cos 2\gamma = 0$$

Continuing from the real part equation:

$$(1) \Rightarrow (2\cos^2\alpha – 1) + (2\cos^2\beta – 1) + (2\cos^2\gamma – 1) = 0$$

$$\Rightarrow 2\cos^2\alpha + 2\cos^2\beta + 2\cos^2\gamma = 1 + 1 + 1$$

$$\Rightarrow 2(\cos^2\alpha + \cos^2\beta + \cos^2\gamma) = 3$$

$$\Rightarrow \cos^2\alpha + \cos^2\beta + \cos^2\gamma = \frac{3}{2}$$

Similarly, equating the imaginary parts, we derive:

$$(1) \Rightarrow (1 – 2\sin^2\alpha) + (1 – 2\sin^2\beta) + (1 – 2\sin^2\gamma) = 0$$

$$\Rightarrow 2\sin^2\alpha + 2\sin^2\beta + 2\sin^2\gamma = 1 + 1 + 1$$

$$\Rightarrow 2(\sin^2\alpha + \sin^2\beta + \sin^2\gamma) = 3$$

$$\Rightarrow \sin^2\alpha + \sin^2\beta + \sin^2\gamma = \frac{3}{2}$$

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LAQ-10 : If cosα+cosβ+cosγ=0=sinα+sinβ+sinγ, then show that
i.cos3α+cos3β+cos3γ=3cos⁡(α+β+γ)
ii.sin3α+sin3β+sin3γ=3sin⁡(α+β+γ)
iii.Cos(α+β)+cos⁡(β+γ)+cos⁡(γ+α)=0

(i) Given that $$\cos\alpha + \cos\beta + \cos\gamma = 0 = \sin\alpha + \sin\beta + \sin\gamma$$

Let $$a = \cos\alpha + i\sin\alpha = \text{cis}\alpha$$ $$b = \cos\beta + i\sin\beta = \text{cis}\beta$$ $$c = \cos\gamma + i\sin\gamma = \text{cis}\gamma$$

Now, $$a + b + c = (\cos\alpha + i\sin\alpha) + (\cos\beta + i\sin\beta) + (\cos\gamma + i\sin\gamma)$$

$$= (\cos\alpha + \cos\beta + \cos\gamma) + i(\sin\alpha + \sin\beta + \sin\gamma) = 0 + i(0) = 0$$

$$a + b + c = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc$$

$$\Rightarrow (\text{cis}\alpha)^3 + (\text{cis}\beta)^3 + (\text{cis}\gamma)^3 = 3\text{cis}\alpha.\text{cis}\beta.\text{cis}\gamma$$

$$\Rightarrow \text{cis}3\alpha + \text{cis}3\beta + \text{cis}3\gamma = 3\text{cis}(\alpha + \beta + \gamma)$$

$$\Rightarrow (\cos3\alpha + i\sin3\alpha) + (\cos3\beta + i\sin3\beta) + (\cos3\gamma + i\sin3\gamma) = 3[\cos(\alpha + \beta + \gamma) + i\sin(\alpha + \beta + \gamma)]$$

$$\Rightarrow (\cos3\alpha + \cos3\beta + \cos3\gamma) + i(\sin3\alpha + \sin3\beta + \sin3\gamma) = 3\cos(\alpha + \beta + \gamma) + i.3\sin(\alpha + \beta + \gamma)$$

(ii) Equating the real parts, we get $$\cos3\alpha + \cos3\beta + \cos3\gamma = 3\cos(\alpha + \beta + \gamma)$$

Hence (i) is proved.

Equating the imaginary parts, we get $$\sin3\alpha + \sin3\beta + \sin3\gamma = 3\sin(\alpha + \beta + \gamma)$$

Hence (ii) is proved.

(iii) Consider, $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{\text{cis}\alpha} + \frac{1}{\text{cis}\beta} + \frac{1}{\text{cis}\gamma}$$

$$= (\cos\alpha – i\sin\alpha) + (\cos\beta – i\sin\beta) + (\cos\gamma – i\sin\gamma)$$

$$= (\cos\alpha + \cos\beta + \cos\gamma) – i(\sin\alpha + \sin\beta + \sin\gamma) = 0 – i(0) = 0$$

$$\Rightarrow \frac{bc + ca + ab}{abc} = 0$$

$$\Rightarrow ab + bc + ca = 0$$

$$\Rightarrow (\text{cis}\alpha)(\text{cis}\beta) + (\text{cis}\beta)(\text{cis}\gamma) + (\text{cis}\gamma)(\text{cis}\alpha) = 0$$

$$\Rightarrow \text{cis}(\alpha + \beta) + \text{cis}(\beta + \gamma) + \text{cis}(\gamma + \alpha) = 0$$

$$\Rightarrow [\cos(\alpha + \beta) + \cos(\beta + \gamma) + \cos(\gamma + \alpha)] + i[\sin(\alpha + \beta) + \sin(\beta + \gamma) + \sin(\gamma + \alpha)] = 0 + i(0) = 0$$

Now, equating the real parts, we get $$\cos(\alpha + \beta) + \cos(\beta + \gamma) + \cos(\gamma + \alpha) = 0$$

Hence (iii) is proved.