How do you divide # (-4+5i) / (2+i) # in trigonometric form?

1 Answer
Shiva Prakash M V
Feb 4, 2018

#(-4+5i)/(2+i)#
in trigonometric form is
#(r1)/(r2)cis(theta1-theta2)#
where,
#r1 = sqrt(((-4)^2+5^2))#
#theta1 =tan^-1(5/(-4))#
and
#r2 = sqrt((2^2+1^2))#
#theta1 =tan^-1(1/2)#

Explanation:

Divide #(-4+5i)/(2+i)#
in trigonometric form
Expressing numerator and denominator as
# z = r cis theta#
For #z1=-4+5i#
#r1 = sqrt(((-4)^2+5^2))#
#theta1 =tan^-1(5/(-4))#
For #z2=2+i#
#r2 = sqrt((2^2+1^2))#
#theta1 =tan^-1(1/2)#
Thus,
#(-4+5i)/(2+i)=(r1cistheta1)/(r2cistheta2)#
#=(r1)/(r2)(cistheta1)/(cistheta2)#
By De-Moivre's theorem,
#(cistheta1)/(cistheta2)=cis(theta1-theta2)#
Now,
#(-4+5i)/(2+i)=(r1)/(r2)cis(theta1-theta2)#

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