How do you add #(1+7i)+(-4-2i)# in trigonometric form?

1 Answer
Jun 25, 2018

Answer:

#color(brown)(=> -3 + 5i#

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+7^2))=sqrt50#
#r_2=sqrt(-4^2+-2^2) =sqrt20#

#theta_1=tan^-1(7/1)~~81.87^@, " I quadrant"#
#theta_2=tan^-1(-2/-4)~~206.57^@, " III quadrant"#

#z_1 + z_2 = sqrt50 cos(81.87) + sqrt20 cos(206.57) + i (sqrt50 sin(81.87)+ sqrt20 sin(206.57))#

#=> 1 -4 + i (7- 2 )#
#color(brown)(=> -3 + 5i#

Proof:

#1 + 7i - 4 - 2i#

#=> -3 + 5i#