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Q: 5     Convert the following in the polar form:

             (ii)    \small \frac{1+3i}{1-2i}

Answers (1)

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Let 
z =\frac{1+3i}{1-2i}

Now, multiply the numerator and denominator by  1+2i
\Rightarrow z =\frac{1+3i}{1-2i} \times \frac{1+2i}{i+2i}= \frac{1+2i+3i-6}{1+4}= \frac{-5+5i}{5}=-1+i
Now,
let 
r\cos\theta = -1 \ \ and \ \ r\sin \theta = 1
On squaring both and then add
r^2(\cos^2\theta+\sin^2\theta)= (-1)^2+1^2
r^2=2
r = \sqrt2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)
Now,
\sqrt2 \cos \theta = -1 \ \ and \ \ \sqrt2\sin \theta = 1
\cos \theta = -\frac{1}{\sqrt2} \ \ and \ \ \sin \theta = \frac{1}{\sqrt2}
Since the value of \cos \theta is negative and  \sin \theta  is positive  this is the case in II quadrant
Therefore,
\theta = \pi - \frac{\pi}{4}= \frac{3\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)
z = r\cos \theta + ir\sin \theta
    =\sqrt2\cos \frac{3\pi}{4} + i\sqrt2\sin \frac{3\pi}{4}
    =\sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right )
Therefore,  the required polar form is

 \sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right ) 

Posted by

Gautam harsolia

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