Question

5-i/2-3i in polar form

Answer 1

$on \: rationalising \: denominator \\ \frac{(5 - i)}{(2 - 3i)} \times \frac{(2 + 3i)}{(2 + 3i)} \\ = \frac{(5(2 + 3i) - i(2 + 3i))}{ {(2 )}^{2} - {(3i)}^{2} } \\ = \frac{(10 + 15i - 2i - 3 {i}^{2}) }{4 - 9( - 1)} \\ = \frac{10 + 13i - 3( - 1)}{4 + 9} \\ = \frac{13 + 13i}{13} = 1 + i = z(say)$

therefore,

$\tan( \alpha ) = \frac{1}{1} = 1 \\ \alpha = \frac{\pi}{4}$

and, |z| =

$\sqrt{ {(1)}^{2} + {(1)}^{2} } = \sqrt{2}$

therefore,

polar form of z=

$= |z| ( \cos( \alpha ) + i \sin( \alpha ) ) \\ = \sqrt{2} ( \cos( \frac{\pi}{4} ) + i \sin( \frac{\pi}{4} ) )$