Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\inline \large \dpi{110}\frac{\left (\cos\frac{\pi}{5}-i\sin \frac{\pi}{5}\right )^2}{\left (\cos\frac{\pi}{5}+i\sin \frac{\pi}{5}\right )^3}$ is gelijk aan	A. $\cos\frac{\pi}{5}-i\sin\frac{\pi}{5}$
	B. $\cos\frac{\pi}{5}-i\sin\frac{\pi}{5}$
	C. i
	D. 1
	E. − 1

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Translation in E N G L I S H

$https://latex.codecogs.com/png.image?\inline \large \dpi{110}\frac{\left (\cos\frac{\pi}{5}-i\sin \frac{\pi}{5}\right )^2}{\left (\cos\frac{\pi}{5}+i\sin \frac{\pi}{5}\right )^3}$ is equal to	A. $\cos\frac{\pi}{5}-i\sin\frac{\pi}{5}$
	B. $\cos\frac{\pi}{5}-i\sin\frac{\pi}{5}$
	C. i
	D. 1
	E. − 1

Oplossing - Solution

We houden rekening met de formule van De Moivre zodat
(cosα + i.sinα)⁻¹ = cos(−α) + i.sin(−α) = cosα − i.sinα en
$\\\frac{\left(\cos\frac{\pi}{5}-i\sin\frac{\pi}{5}\right)^2}{\left(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\right)^3}=\frac{\left(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\right)^{-2}}{\left(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\right)^3}=\frac{1}{\left(\cos\frac{\pi}{5}-i\sin\frac{\pi}{5}\right)^5}\\=\frac{1}{\cos\pi+i\sin\pi}=\frac{1}{-1}=-1$