Gujarat 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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