How do you add #(-5+5i)+(7+i)# in trigonometric form?

1 Answer
Jun 25, 2018

Answer:

#color(cyan)(=> 2 + 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(-5^2+ 5^2))=sqrt 50#
#r_2=sqrt(7^2+ 1^2) =sqrt 50#

#theta_1=tan^-1(5/-5)~~ 135^@, " II quadrant"#
#theta_2=tan^-1(1/ 7)~~ 8.13^@, " I quadrant"#

#z_1 + z_2 = sqrt 50 cos(135) + sqrt 50 cos(8.13) + i (sqrt 50 sin 135 + sqrt 50 sin 8.13)#

#=> -5+ 7 + i (5 + 1 )#

#color(cyan)(=> 2 + 6 i#