How do you add #(2+9i)+(5-7i)# in trigonometric form?

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Answer 1 · 2018-05-21T14:58:20

#(2+9i)+(5-7i)=sqrt53(cos(0.279)-sin(0.279)i)#

We add two complex numbers #a+bi# and #c+di# as follows:

#(a+bi)+(c+di):=(a+c)+(b+d)i#

So #(2+9i)+(5-7i)=7-2i#. Now, to convert to trigonometric, we use the following identity:

#(a+bi)=r(cos(theta)+sin(theta)i)#

where #r=sqrt(a^2+b^2)# and #theta# satisifies #costheta=a/r,sintheta=b/r#

So, for #7-2i#, #r =sqrt(49+4)=sqrt53# and #costheta=7"/"sqrt53,sintheta=-2"/"sqrt53# and so #theta=arctan(-2"/"7)approx-0.279#

Finally, #7-2i=sqrt53(cos(-0.279)+sin(-0.279)i)=sqrt53(cos(0.279)-sin(0.279)i)#

Answer 2 · 2018-07-09T11:53:02

#color(violet)(=> 7 + 2 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(2^2+ 9^2))=sqrt 85#
#r_2=sqrt(5^2+ -7^2) =sqrt 74#

#theta_1=tan^-1(9 / 2)~~ 77.47^@, " I quadrant"#
#theta_2=tan^-1(-7/ 5)~~ 305.54^@, " IV quadrant"#

#z_1 + z_2 = sqrt 85 cos(77.47) + sqrt 74 cos(305.54) + i (sqrt 85 sin 77.47 + sqrt 74 sin 305.54)#

#=> 2 + 5 + i (9 - 7 )#

#color(violet)(=> 7 + 2 i#