How do you add #(-1+8i)+(4-2i)# in trigonometric form?

1 Answer
Jun 25, 2018

Answer:

#color(indigo)(=> 3+ 6 i#

Explanation:

#z= a+bi= r (costheta+isintheta)#

#r=sqrt(a^2+b^2), " " theta=tan^-1(b/a)#

#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))#

#r_1=sqrt(-1^2+ 8^2))=sqrt 65#
#r_2=sqrt(-2^2+ 4^2) =sqrt 20#

#theta_1=tan^-1(8 / -1)~~ 97.13^@, " II quadrant"#
#theta_2=tan^-1(-2/ 4)~~ 333.43^@, " IV quadrant"#

#z_1 + z_2 = sqrt 65 cos(97.13) + sqrt 20 cos(333.43) + i (sqrt 65 sin 97.13 + sqrt 20 sin 333.43)#

#=> -1 + 4 + i (8 - 2 )#

#color(indigo)(=> 3+ 6 i#