How do you multiply # (1+3i)(1-3i) # in trigonometric form?
1 Answer
Mar 27, 2016
#(1 + 3i)(1 - 3i) = 10#
Explanation:
#abs(1+3i) = abs(1-3i) = sqrt(1^2+3^2) = sqrt(10)#
#1 + 3i = sqrt(10) cis (arctan(3))#
#1 - 3i = sqrt(10) cis (arctan(-3)) = sqrt(10) cis (-arctan(3))#
So:
#(1 + 3i)(1 - 3i)#
#= (sqrt(10) cis (arctan(3)))(sqrt(10) cis (-arctan(3)))#
#= (sqrt(10))^2 cis (arctan(3) - arctan(3))#
#= 10 cis(0)#
#= 10#
