How do you divide #( i+9) / (i -2 )# in trigonometric form?

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Answer 1 · 2017-10-31T10:59:42

# (9+i)/(-2+i) = 4.05 (cos 3.72+isin3.72) #

#Z= (9+i)/(-2+i) = ((9+i)(-2-i))/((-2+i)(-2-i)) #

# =(-18-11i-i^2)/((-2)^2-i^2) = (-17-11i)/(4+1) = -17/5-11/5i#

#= -3.4 -2.2i ; [i^2=-1]#

Modulus #|Z|=r=sqrt((-3.4)^2+ (-2.2)^2) =4.05 # ;

#tan alpha =b/a= (-2.2)/-3.4 = 0.647 :. alpha =tan^-1(0.647) = 0.574#

#theta# is on #3rd# quadrant # :. theta= alpha+pi=0.574+pi= 3.72#;

#theta# expressed in radian.

Argument : # theta =3.72 :. # In trigonometric form expressed as

#r(cos theta=isintheta) = 4.05(cos 3.72+isin3.72) :.#

# (9+i)/(-2+i) = 4.05(cos 3.72+isin3.72) # [Ans]