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

1 Answer
Binayaka C.
Oct 31, 2017

(9+i)/(-2+i) = 4.05 (cos 3.72+isin3.72) 9+i−2+i=4.05(cos3.72+isin3.72)

Explanation:

Z= (9+i)/(-2+i) = ((9+i)(-2-i))/((-2+i)(-2-i)) 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=−18−11i−i2(−2)2−i2=−17−11i4+1=−175−115i

= -3.4 -2.2i ; [i^2=-1]=−3.4−2.2i;[i2=−1]

Modulus |Z|=r=sqrt((-3.4)^2+ (-2.2)^2) =4.05 |Z|=r=√(−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]

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