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

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Answer 1 · 2016-10-24T11:49:06

#= sqrt(181) *(cos(0.83798)+i sin(0.83798))#

It is usually easier to simplify cartesian form and then convert to trigonometric afterwards,

#(7+9i)+(2+i)#

#9+10i#

now we convert to trigonometric.

#9+10i = r*cis(theta)#

#r=sqrt(9^2+10^2)#

#r=sqrt(181)#

#theta=tan^-1(10/9)#

#theta=0.83798#

giving us,

#(7+9i)+(2+i) = sqrt(181) *cis(0.83798)#

or in expanded form,

#= sqrt(181) *(cos(0.83798)+i sin(0.83798))#