Convert each of the complex numbers in the polar form: 

Q : 6        -1-i

Answers (1)
G Gautam harsolia

Given problem is
z=-1-i
Now, let 
r\cos \theta = -1 \ \ \ and \ \ \ r\sin \theta = -1
Square and add both the sides 
r^2(\cos^2\theta +\sin^2\theta)= (-1)^2+(-1)^2                                                     (\because \cos^2\theta +\sin^2\theta = 1)
r^2= 1+1
r^2 =2
r= \sqrt2                                                                                                                         (\because r > 0)
Therefore, the modulus is \sqrt2
Now, 
\sqrt2\cos \theta = -1 \ \ \ and \ \ \ \sqrt2\sin \theta = -1
\cos \theta = -\frac{1}{\sqrt2} \ \ \ and \ \ \ \sin \theta =-\frac{1}{\sqrt2}
Since values of both  \cos \theta and \sin \theta is negative  and  we know that this is the case in  III quadrant
Therefore,
\theta =-\left ( \pi - \frac{\pi}{4} \right )= -\frac{3\pi}{4}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ III \ quadrant)
Therefore,
-1-i= r\cos \theta +ir\sin \theta
               = \sqrt2\cos \left ( -\frac{3\pi}{4} \right ) +i\sqrt2\sin \left (- \frac{3\pi}{4} \right )
               = \sqrt2\left ( \cos \left ( -\frac{3\pi}{4} \right ) +i\sin \left ( -\frac{3\pi}{4} \right ) \right )

Therefore, the required polar form is   \sqrt2\left ( \cos \left (- \frac{3\pi}{4} \right ) +i\sin \left (- \frac{3\pi}{4} \right ) \right )

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