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 )
