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Convert each of the complex numbers in the polar form: 

Q : 8    i

Answers (1)

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Given problem is
z = i
Now, let 
r\cos \theta = 0 \ \ \ and \ \ \ r\sin \theta = 1
Square and add both the sides 
r^2(\cos^2\theta +\sin^2\theta)= (0)^2+(1)^2                                                     (\because \cos^2\theta +\sin^2\theta = 1)
r^2= 0+1
r^2 =1
r= 1                                                                                                                         (\because r > 0)
Therefore, the modulus is 1
Now, 
1\cos \theta =0 \ \ \ and \ \ \ 1\sin \theta = 1
\cos \theta =0\ \ \ and \ \ \ \sin \theta =1
Since values of Both \cos \theta  and \sin \theta is Positive  and  we know that this is the case in  I quadrant
Therefore,
\theta =\frac{\pi}{2}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ I \ quadrant)
Therefore,
i= r\cos \theta +ir\sin \theta
   = 1\cos \left (\frac{\pi}{2} \right ) +i1\sin \left (\frac{\pi}{2} \right )
   = \cos \frac{\pi}{2} +i\sin\frac{\pi}{2}

Therefore, the required polar form is   \cos \frac{\pi}{2} +i\sin\frac{\pi}{2}

Posted by

Gautam harsolia

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