How do you multiply # (5-i)(6-4i) # in trigonometric form?

1 Answer
Yonas Yohannes
Apr 10, 2016

#C= sqrt(26)*sqrt(52)/_(349^0+326^0)#
#C~~36.8/_315^0#

Explanation:

Given: # C_1= (5-i), C_2=(6-4i) #

Required: The product of #C_1*C_2# in trigonometric form

Solution Strategy:
1) Convert the Phasors (another name for complex numbers) to their polar or trig form. That is given by:
#C_1 = |C_1| (cos theta +isintheta)#
where: #|C_1|= sqrt(5^2+1^2)=sqrt(26); theta= tan^-1 (-1/5)=349#

#C_2 = |C_2| (cos alpha+isinalpha)#
#|C_2|= sqrt(6^2+4^2)=sqrt(52); theta= tan^-1 (-4/6)=326#

2) Product: #C_1*C_2= sqrt(26)/_349^o*sqrt(52)/_326^o#
# = sqrt(26)*sqrt(52)/_(349^0+326^0)#

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