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

1 Answer
Monzur R.
Nov 8, 2017

#(5+5i)+(-3+i)=2sqrt10(cos(5/4)+isin(5/4))#

Explanation:

#(5+5i)+(-3+i)=(5-3)+(5i+i)=2+6i#

Any complex number of the form #a+bi, a,b in RR# may be written in its mod-arg form #r(cosvartheta+isinvartheta)# where #r=sqrt(a^2+b^2)# and #vartheta =arctan (b"/"a)#

#r=sqrt(2^2+6^2)=sqrt(40)=2sqrt10#

#vartheta=arctan (6/2)=arctan3approx5/4#

#therefore 2+6i = 2sqrt10(cos(5/4)+isin(5/4))#

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