How do you multiply # (-7+3i)(1+3i) # in trigonometric form?

1 Answer
May 14, 2018

#(-7+ 3 i)(1+ 3 i) =24.08 ( cos (3.99) + i sin(3.99) = (-16-18 i) #

Explanation:

Let #Z=a+i b ; Z=-7+ 3 i ; a=-7 ,b = 3# ;

#Z=-7+ 3 i# is in #2# nd quadrant.

Modulus #|Z|=sqrt(a^2+b^2)=(sqrt((-7)^2+ 3^2)) =sqrt 58 #

# tan alpha =|b/a|= 3/7 or alpha =tan^-1(3/7) ~~ 0.4049#

#theta# is on #2# nd quadrant # :. theta=pi-0.4049#

# :. theta~~ 2.7367#. Argument , # theta ~~2.7367:. #

In trigonometric form expressed as

#r(cos theta+isintheta) = sqrt58(cos 2.74+i sin 2.74) #

#Z=1+ 3 i# is in #1# st quadrant.

Modulus #|Z|=sqrt(a^2+b^2)=(sqrt(1^2+ 3^2)) =sqrt 10 #

# tan alpha =|b/a|= 3/1 or alpha =tan^-1(3) ~~ 1.249#

#theta# is on #1# st quadrant # :. theta=1.249#

Argument , # theta ~~1.249:. #

In trigonometric form expressed as

#r(cos theta+isintheta) = sqrt 10 (cos 1.25+i sin 1.25) #

#(-7+ 3 i)(1+ 3 i) = #

# sqrt58(cos 2.74+i sin 2.74) * sqrt 10 (cos 1.25+i sin 1.25)~~ #

#sqrt58 * sqrt 10 ( cos (2.74+1.25) + i sin(2.74+1.25) ~~#

#24.08 ( cos (3.99) + i sin(3.99) = (-16-18 i)# [Ans]