How do you add #(8+4i)# and #(3-3i)# in trigonometric form?

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Answer 1 · 2018-06-25T09:59:49

#color(crimson)(=> 11+ 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(4^2+ 8^2))=sqrt 80#
#r_2=sqrt(3^2+ -3^2) =sqrt 18#

#theta_1=tan^-1(4 / 8)~~ 26.57^@, " I quadrant"#
#theta_2=tan^-1(-3/ 3)~~ 315^@, " IV quadrant"#

#z_1 + z_2 = sqrt 80 cos(26.57) + sqrt 18 cos(315) + i (sqrt 80 sin 26.57 + sqrt 18 sin 315)#

#=> 8 + 3 + i (4 - 3 )#

#color(crimson)(=> 11+ i#