How do you add (-2-3i)+(6-9i)(23i)+(69i) in trigonometric form?

1 Answer
Jun 25, 2018

color(blue)(=> 4 - 12 i412i

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(-2^2+ -3^2))=sqrt 13r1=22+32)=13
r_2=sqrt(6^2+ -9^2) =sqrt 117r2=62+92=117

theta_1=tan^-1(-3/-2)~~ 236.31^@, " III quadrant"θ1=tan1(32)236.31, III quadrant
theta_2=tan^-1(-9/ 6)~~ 303.69^@, " IV quadrant"θ2=tan1(96)303.69, IV quadrant

z_1 + z_2 = sqrt 13 cos(236.31) + sqrt 117 cos(303.69) + i (sqrt 13 sin 236.31 + sqrt 117 sin 303.69)z1+z2=13cos(236.31)+117cos(303.69)+i(13sin236.31+117sin303.69)

=> -2+ 6 + i (-3 - 9 )2+6+i(39)

color(blue)(=> 4 - 12 i412i