How do you multiply # (3-5i)(7-8i) # in trigonometric form?

Question

1606 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-17T12:02:47

# (3-5i)(7-8i) = 61.98(cos 4.40+isin4.40) #

#Z= (3-5i)(7-8i) = (21-24i-35i+40i^2)#

# =21-59i-40 = -19-59i [i^2= -1]#

Modulus #|Z|=r=sqrt((-19)^2+ (-59)^2) =61.98 # ;

#tan alpha =b/a= (-59)/-19 = 3.105 :. alpha =tan^-1(3.105) = 1.259# or

#theta# is on #3rd# quadrant # :. theta=1.259+pi= 4.40 # rad

Argument : # theta =4.40 :. # In trigonometric form expressed as

#r(cos theta+isintheta) = 61.98(cos 4.40+isin4.40) :.#

# (3-5i)(7-8i) = 61.98(cos 4.40+isin4.40) # [Ans]