6 Most VSAQ’s of De Moivre’s Theorem Chapter in Inter 2nd Year Maths-2A (TS/AP)

2 Marks

VSAQ-1 : If A,B,C are angles of a triangle, x=cisA, y=cisB, z=cisC, then find the value of xyz

In $$\Delta ABC$$ we have, $$A + B + C = 180^\circ$$

$$xyz = \text{cis}A \cdot \text{cis}B \cdot \text{cis}C = \text{cis}(A+B+C) = \text{cis}180^\circ$$

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

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VSAQ-2 : If x=cosθ+isinθ then find x⁶+1/x⁶

$$x^6 = (\cos\theta + i\sin\theta)^6 = \cos6\theta + i\sin6\theta$$

$$\Rightarrow \frac{1}{x^6} = \cos6\theta – i\sin6\theta$$

$$x^6 + \frac{1}{x^6} = 2\cos6\theta$$

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VSAQ-3 : If 1,ω,ω² are the cube roots of unity, show that (1-ω+ω²)⁶+(1+ω-ω²)⁶=128

We know that $$1 + \omega + \omega^2 = 0$$

Hence, we get $$\omega + \omega^2 = -1$$ $$1 + \omega = -\omega^2$$

$$(1 – \omega + \omega^2)^6 + (1 + \omega – \omega^2)^6 = (1 + \omega^2 – \omega)^6 + (1 + \omega – \omega^2)^6$$

$$= (-\omega – \omega)^6 + (-\omega^2 – \omega^2)^6 = (-2\omega)^6 + (-2\omega^2)^6 = 2^6\omega^6 + 2^6\omega^{12} = 64(\omega^3)^2 + 64(\omega^3)^4$$

$$= 64(1)^2 + 64(1)^4 = 64 + 64 = 128$$

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VSAQ-4 : If 1,ω,ω² are the cube roots of unity then prove that (2-ω)(2-ω²)(2-ω¹⁰)(2-ω¹¹)=49

$$\omega^{10} = (\omega^9)\omega = (\omega^3)^3\omega = 1(\omega) = \omega$$

$$\omega^{11} = (\omega^{10})\omega = (\omega)\omega = \omega^2$$

$$(2 – \omega)(2 – \omega^2)(2 – \omega^{10})(2 – \omega^{11})$$

$$= \left[4 – 2(\omega^2 + \omega) + \omega^3\right]^2 = \left[4 – 2(\omega^2 + \omega) + 1\right]^2 = \left[5 – 2(\omega^2 + \omega)\right]^2 = \left[5 – 2(-1)\right]^2$$

$$= (5 + 2)^2 = 7^2 = 49$$

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VSAQ-5 : Find the value of (1-i)⁸

$$1 – i = \sqrt{2}\left[\frac{1}{\sqrt{2}} – i \frac{1}{\sqrt{2}}\right]$$

$$= \sqrt{2}[\cos \frac{\pi}{4} – i\sin \frac{\pi}{4}]$$

$$(1 – i)^8 = [\sqrt{2}(\cos \frac{\pi}{4} – i\sin \frac{\pi}{4})]^8$$

$$= (\sqrt{2})^8[\cos 8 \frac{\pi}{4} – i\sin 8 \frac{\pi}{4}]$$

$$= 2^4(\cos 2\pi – i\sin 2\pi) = 16[1 – i(0)] = 16$$

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VSAQ-6 : Find the value of (1+i)¹⁶

$$1 + i = \sqrt{2}\left[\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right]$$

$$= \sqrt{2}[\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}]$$

$$(1 + i)^{16} = [\sqrt{2}(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4})]^{16}$$

$$= (\sqrt{2})^{16}[\cos 16 \frac{\pi}{4} + i\sin 16 \frac{\pi}{4}]$$

$$= 2^8(\cos 4\pi + i\sin 4\pi) = 256[1 + i(0)] = 256$$