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Complex Numbers and Quadratic Equations — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers and Quadratic Equations2 Marks
Question
Convert the complex number $\frac {-16}{1+\sqrt3i}$ into polar form.
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Answer

Given complex number $\frac {-16}{1+\sqrt3i}$ convert the complex number in $x +iy$ form
$= \frac {-16}{1+\sqrt3i} \times \frac {1-\sqrt3i}{1-\sqrt3i}$
$= \frac {-16(1-\sqrt3i)}{1-(\sqrt3i)^2} = \frac {-16(1-\sqrt3i)}{1+3}$
$= -4 (1 - \sqrt3i) = -4 + 4\sqrt3i$
Let $-4 = r\; \cos \;\theta, 4\sqrt3 = r\; \sin \;\theta$
By squaring and adding, we get
$16 + 48 = r^2(\cos^2 \theta + \sin^2 \theta)$
which gives $r^2 = 64,$ i.e., $r = 8$
Hence $\cos \theta = - \frac 12, \sin \theta = \frac {\sqrt3}2$
$\theta = \pi - \frac {\pi}3 = \frac {2\pi}{3}$
Thus, the required polar form is $r(\cos \theta  + i \sin \theta ) =  8\left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)$

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