How do you add #(2+i)+(5+3i)# in trigonometric form?

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Answer 1 · 2016-03-31T02:32:03

#7+4i=sqrt65[cos(tan^-1 (4/7))+i sin (tan^-1 (4/7))]#

From the given
#(2+i)+(5+3i)#

#(2+5)+(i+3i)#

#7+4i#

in Trigonometric Form

#a+bi=sqrt(a^2+b^2)[cos (tan^-1 (b/a))+i sin (tan^-1 (b/a))]#

#7+4i=sqrt65[cos(tan^-1 (4/7))+i sin (tan^-1 (4/7))]#

God bless....I hope the explanation is useful.