How do you add (-8+9i)+(4+6i)(8+9i)+(4+6i) in trigonometric form?

1 Answer
Jun 25, 2018

color(crimson)(=> -4 + 15 i4+15i

Explanation:

z= a+bi= r (costheta+isintheta)z=a+bi=r(cosθ+isinθ)

r=sqrt(a^2+b^2), " " theta=tan^-1(b/a)r=a2+b2, θ=tan1(ba)

r_1(cos(theta_1)+isin(theta_2))+r_2(cos(theta_2)+isin(theta_2))=r_1cos(theta_1)+r_2cos(theta_2)+i(r_1sin(theta_1)+r_2sin(theta_2))r1(cos(θ1)+isin(θ2))+r2(cos(θ2)+isin(θ2))=r1cos(θ1)+r2cos(θ2)+i(r1sin(θ1)+r2sin(θ2))

r_1=sqrt(-8^2+ 9^2))=sqrt 145r1=82+92)=145
r_2=sqrt(4^2+ 6^2) =sqrt 52r2=42+62=52

theta_1=tan^-1(9 / -8)~~ 131.63^@, " II quadrant"θ1=tan1(98)131.63, II quadrant
theta_2=tan^-1(6/ 4)~~ 56.31^@, " I quadrant"θ2=tan1(64)56.31, I quadrant

z_1 + z_2 = sqrt 145 cos(131.63) + sqrt 52 cos(56.31) + i (sqrt 145 sin 131.63 + sqrt 52 sin 56.31)z1+z2=145cos(131.63)+52cos(56.31)+i(145sin131.63+52sin56.31)

=> -8 + 4 + i (9 + 6 )8+4+i(9+6)

color(crimson)(=> -4 + 15 i4+15i