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Convert each of the complex numbers in the polar form: (6) -3

Convert each of the complex numbers in the polar form: 

Q: 6        -3

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Given problem is
z=-3
Now, let 
r\cos \theta = -3 \ \ \ and \ \ \ r\sin \theta = 0
Square and add both the sides 
r^2(\cos^2\theta +\sin^2\theta)= (-3)^2+(0)^2                                                     (\because \cos^2\theta +\sin^2\theta = 1)
r^2= 9+0
r^2 =9
r= 3                                                                                                                         (\because r > 0)
Therefore, the modulus is 3
Now, 
3\cos \theta =- 3 \ \ \ and \ \ \ 3\sin \theta = 0
\cos \theta = -1\ \ \ and \ \ \ \sin \theta =0
Since values of  \cos \theta is negative and \sin \theta is Positive  and  we know that this is the case in  II quadrant
Therefore,
\theta =\pi\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)
Therefore,
-3= r\cos \theta +ir\sin \theta
   = 3\cos \left (\pi \right ) +i3\sin \left (\pi \right )
   = 3\left ( \cos \pi +i\sin\pi \right )

Therefore, the required polar form is   3\left ( \cos \pi +i\sin\pi \right )

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