How do you add #(7+4i)# and #(2+6i)# in trigonometric form?

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Answer 1 · 2018-01-29T08:59:28

#9+10i#

If we have a complex number such that #z=a+bi#, then #z=r(costheta+isintheta)#, where:

For #z_1=7+4i# :
#r=sqrt(7^2+4^2)=sqrt(49+16)=sqrt(65)#
#theta=tan^(-1)(4/7)~~29.7#
#=sqrt(65)(cos(29.7)+isin(29.7))#

For #z_2=2+6i# :
#r=sqrt(2^2+6^2)=sqrt(4+36)=sqrt(40)#
#theta=tan^(-1)(6/2)~~71.6#
#=sqrt(40)(cos(71.6)+isin(71.6))#

For #z_1+z_2# :
#z_1+z_2=sqrt(65)cos(29.7)+isqrt(65)sin(29.7)+sqrt(40)cos(71.6)+isqrt(40)sin(71.6)#

#color(white)(z_1+z_2)=sqrt(65)cos(29.7)+sqrt(40)cos(71.6)+i(sqrt(65)sin(29.7)+sqrt(40)sin(71.6))#

#color(white)(z_1+z_2)=8.999470954+9.995734316i##

#color(white)(z_1+z_2)~~9+10i#

Proof:
#7+4i+2+6i=(7+2)+i(4+6)=9+10i#