How do you add (-9+i)+(-1+3i)(9+i)+(1+3i) in trigonometric form?

1 Answer
Jul 9, 2018

color(crimson)(=> -10+ 4i10+4i

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(-9^2+ 1^2))=sqrt 82r1=92+12)=82
r_2=sqrt(-1^2+ 3^2) =sqrt 10r2=12+32=10

theta_1=tan^-1(1 / -9)~~ 173.66^@, " II quadrant"θ1=tan1(19)173.66, II quadrant
theta_2=tan^-1(3/ -1)~~ 108.43^@, " II quadrant"θ2=tan1(31)108.43, II quadrant

z_1 + z_2 = sqrt 82 cos(173.66) + sqrt 10 cos(108.43) + i (sqrt 82 sin 173.66 + sqrt 10 sin 108.43)z1+z2=82cos(173.66)+10cos(108.43)+i(82sin173.66+10sin108.43)

=> -9 - 1 + i (1 + 3 )91+i(1+3)

color(crimson)(=> -10+ 4i10+4i