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

1 Answer
Aug 14, 2018

#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 )