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How do you add #(2-5i)# and #(-2-2i)# in trigonometric form?

1 Answer
Jul 9, 2018

Answer:

#color(red)(=> - 7i#

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(2^2+ -5^2))=sqrt 29#
#r_2=sqrt(-2^2+ -2^2) =sqrt 8#

#theta_1=tan^-1(-5 / 2)~~ 291.8^@, " IV quadrant"#
#theta_2=tan^-1(-2/ -2)~~ 225^@, " III quadrant"#

#z_1 + z_2 = sqrt 29 cos(291.8) + sqrt 8 cos(225) + i (sqrt 29 sin 291.8 + sqrt 8 sin 225)#

#=> 2 - 2 + i (-5 - 2 )#

#color(red)(=> - 7i#