Browse Questions

# Convert the given complex number in polar form :$\;-1+i\;$

$(a)\;\sqrt{2}\;[cos (\large\frac{\pi}{4})+i \;sin(\large\frac{\pi}{4})]\qquad(b)\;\sqrt{2}\;[cos (\large\frac{3\pi}{4})+i \;sin(\large\frac{3\pi}{4})]\qquad(c)\;\sqrt{2}\;[cos (\large\frac{5\pi}{4})+i \;sin(\large\frac{5\pi}{4})]\qquad(d)\;0$

Answer : $\;\sqrt{2}\;[cos (\large\frac{3\pi}{4})+i \;sin(\large\frac{3\pi}{4})]$
Explanation :
$\;-1+i\;$ Let $\;r cos \theta =-1 \;$ and $\;r sin \theta = 1$
on squaring and adding , we obtain
$(r cos \theta)^{2}+(r sin \theta)^{2} = (-1)^{2}+(1)^{2}$
$r^{2}(cos^{2} \theta + sin^{2} \theta) = 1+ 1$
$r^{2} = 2 \qquad [cos^{2} \theta + sin^{2} \theta=1]$
$r =\sqrt{2} \qquad [conventionally , r > 0 ]$
Therefore , $\; \sqrt{2} cos \theta =- 1 \;$ and $\; \sqrt{2} sin \theta = -1$
$cos \theta = \large\frac{-1}{\sqrt{2}}\;$ and $\;sin \theta = \large\frac{1}{\sqrt{2}}$
$\theta =(\pi - \large\frac{\pi}{4})= \large\frac{3\pi}{4}\qquad [\theta \;lies\; in\;the\;2^{nd} \; quadrant]$
It can be written
$-1+i = r cos \theta + i r sin \theta$
$=\sqrt{2}\;[cos (\large\frac{3\pi}{4})+i \;sin(\large\frac{3\pi}{4})]\;$ this is the required polar form .