How do you divide #( 4i+1) / (-8i +5 )# in trigonometric form?

1 Answer
John D.
Mar 30, 2017

#0.315i - 0.303#

Explanation:

First, convert both the numerator and denominator into polar form.

Converting #4i+1# to polar:

#r = sqrt(4^2 + 1^2) = sqrt17#
#theta = tan^-1(4/1) = 75.96^@#

#4i + 1 = sqrt17 color(white)"-" angle color(white)"."75.96^@#

Converting #-8i + 5# to polar:

#r = sqrt((-8)^2 + 5^2) = sqrt89#
#theta = tan^-1(-8/5) = -57.99^@#

#-8i + 5 = sqrt89 color(white)"-" angle color(white)"."-57.99^@#

Dividing in polar form:

#(sqrt17 color(white)"-" angle color(white)"."75.96^@) / ( sqrt89 color(white)"-" angle color(white)"."-57.99^@) = sqrt17/sqrt89 color(white)"-" angle color(white)"-" (75.96-(-57.99))^@#

# = sqrt17/sqrt89 color(white)"-" angle color(white)"." 133.95^@#

Now to convert back to rectangular:

#Re = r cos theta = sqrt17 / sqrt89 cos133.95^@ = -0.303#

#Im = ri sin theta = sqrt17/sqrt89 i color(white)"." sin133.95^@ = 0.315i#

So the final answer is #0.315i - 0.303#

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