Question

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

Answer 1

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

Explanation:

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 )