How do you add #(2-3i)# and #(12-2i)# in trigonometric form?
1 Answer
May 17, 2016
Explanation:
A complex number z = x +iy can be expressed in trig. form as shown.
#z=x+iy=r(costheta+isintheta)" where"#
#•r=sqrt(x^2+y^2)" and " theta=tan^-1(y/x)# Now to get this sum in trig form we have to add the numbers together and then convert to trig.
#rArr(2-3i)+(12-2i)=14-5i# Using x = 14 and y = -5 , convert to trig form.
#rArrr=sqrt(14^2+(-5)^2)=sqrt221" does not simplify further"# and
#theta=tan^-1(-5/14)≈-0.343" radians"#
#rArr14-5i=sqrt221(cos(-0.343)+isin(-0.343))# using
#cos(-theta)=costheta" and "sin(-theta)=-sintheta# we can also express in trig form as
#14-5i=sqrt221(cos(0.343)-isin(0.343))#
