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Convert each of the complex numbers in the polar form: (3) 1 - i

Convert each of the complex numbers in the polar form: 

Q : 3        1-i

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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   \sin \theta  is negative and  value \cos \theta is positive and  we know that this is the case in the IV quadrant
Therefore,
\theta = -\frac{\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ IV \ quadrant)
Therefore,
1-i= r\cos \theta +ir\sin \theta
            = \sqrt2\cos \left ( -\frac{\pi}{4} \right ) +i\sqrt2\sin \left ( -\frac{\pi}{4} \right )
            = \sqrt2\left ( \cos \left ( -\frac{\pi}{4} \right ) +i\sin \left ( -\frac{\pi}{4} \right ) \right )

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

 

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