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Answer 1 · 2017-11-07T12:13:16

Modulus #|Z|=sqrt13#, Argument # theta= 56.31^0 # and
polar form : #sqrt13(cos56.31+isin 56.31)#

Let #Z=a+ib ; Z=2+3i ; a=2 ,b =3# ;

#Z=2+3i# is in 1st quadrant

Modulus #|Z|=sqrt(a^2+b^2)=sqrt(2^2+ 3^2) =sqrt13=3.61(2dp)#

# tan theta =b/a= 3/2 or tan theta =1.5 #

Argument # theta =tan^-1 (1.5) = 56.31^0(2dp) #.

In polar form expressed as #|Z|*(costheta+isin theta)#

#sqrt13(cos56.31+isin 56.31)# [Ans]