How do you simplify #(8+6i)-(2+3i)#?

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Answer 1 · 2016-10-23T08:25:23

#3*sqrt(5)*(cos(0.464)+isin(0.464))#

First simplify Cartesian form,
#(8+6i)-(2+3i)= 8-2 +6i-3i#
# =6+3i #

now to convert to trigonometric form,

#6+3i = r*cis(theta)#

where,
#r=sqrt(6^2+3^2)#
#r= sqrt(45)# (which can be further simplified to #3*sqrt(5)#)

and,

#theta=tan^-1(3/6)#

#theta = 0.464# (3dp, radians)

so we can express #(8+6i)-(2+3i)# as,

#3*sqrt(5)*cis(0.464)#

or in expanded version,
#3*sqrt(5)*(cos(0.464)+isin(0.464))#