How do you divide $(9+2i) / (5+i)$ in trigonometric form?

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Answer 1 · 2018-08-14T16:41:53

$sqrt(85/26) ( cos 1.22^o + i sin 1.22^o )$

Use $e^(i theta ) = cos theta + i sin theta$

$( 9 + 2i )/( 5 + i )$

= $( a e^(i alpha) )/ ( b e^(i beta )) = (a/b) (e^(i(alpha - beta )))$ ,

#= a/b ( cos ( alpha - beta) + i sin ( alpha - beta )

where

$a = sqrt( 9^2 + 2^2 ) = sqrt 85$ ,

$b = sqrt ( 5^2 + 1 ) = sqrt26$ ,

$alpha = arccos( 9/a)$ and

$beta = arccos(5/b)$ .

Answer;

$( 9 + 2i )/( 5 + i )$

$= sqrt(85/26)( cos ( arccos (9/sqrt85) - arccos ( 5/sqrt26))$

$+ i sin ( arccos (9/sqrt85) - arccos ( 5/sqrt26) )$

$= sqrt(85/26) ( cos ( 12.53^o - 11.31^o)$

$+i sin ( 12.53^o - 11.31^o) )$

$= sqrt(85/26) ( cos 1.22^o + i sin 1.22^o )$

This is very very close to the value

1/26 ( 47 + i )

How do you divide (9+2i) / (5+i) (9+2i) / (5+i) in trigonometric form?