How do you add #(9-7i)+(1+i)# in trigonometric form?

1 Answer
Jul 9, 2018

Answer:

#color(chocolate)(=> 10 - 6i#

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(9^2+ -7^2))=sqrt 130#
#r_2=sqrt(1^2+ 1^2) =sqrt 2#

#theta_1=tan^-1(-7 / 9)~~ 322.13^@, " IV quadrant"#
#theta_2=tan^-1(1/ 1)~~ 45^@, " I quadrant"#

#z_1 + z_2 = sqrt 130 cos(322.13) + sqrt 2 cos(45) + i (sqrt 130 sin 322.13 + sqrt 2 sin 45)#

#=> 9 + 1 + i (-7 + 1 )#

#color(chocolate)(=> 10 - 6i#