How do you add #(8-2i)+(7-3i)# in trigonometric form?

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Answer 1 · 2018-06-25T08:41:12

#color(cyan)(=> 15 - 5 i#

#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(8^2+-2^2))=sqrt 68#
#r_2=sqrt(7^2+-3^2) =sqrt 58#

#theta_1=tan^-1(-2/8)~~ 345.96^@, " IV quadrant"#
#theta_2=tan^-1(-3/ 7)~~ 336.8^@, " IV quadrant"#

#z_1 + z_2 = sqrt 68 cos(345.96) + sqrt 58 cos(336.8) + i (sqrt 68 sin 345.96 + sqrt 58 sin 336.8)#

#=> 8 + 7 + i (-2 - 3 )#

#color(cyan)(=> 15 - 5 i#