Convert the following into polar form: frac{1 | vedclass

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(A) Given $z = \frac{1+3i}{1-2i}$.Multiply numerator and denominator by the conjugate $(1+2i)$:$z = \frac{(1+3i)(1+2i)}{(1-2i)(1+2i)} = \frac{1+2i+3i+

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$\sqrt{2}\left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right)$

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(A) Given $z = \frac{1+3i}{1-2i}$.Multiply numerator and denominator by the conjugate $(1+2i)$:$z = \frac{(1+3i)(1+2i)}{(1-2i)(1+2i)} = \frac{1+2i+3i+6i^2}{1^2+2^2} = \frac{1+5i-6}{5} = \frac{-5+5i}{5} = -1+i$.Let $z = r(\cos 	heta + i \sin 	heta)$.Here,$r = |z| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1+1} = \sqrt{2}$.Since $z = -1+i$ lies in the second quadrant,$\cos 	heta = \frac{-1}{\sqrt{2}}$ and $\sin 	heta = \frac{1}{\sqrt{2}}$.The reference angle is $\alpha = \frac{\pi}{4}$.Since it is in the second quadrant,$	heta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.Thus,the polar form is $\sqrt{2}\left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}ight)$.

Convert the following into polar form: $\frac{1+3i}{1-2i}$

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