(6−6i)/(–√3 + i) in exponential form?

1 Answer
Feb 15, 2018

The answer is #=3sqrt2e^(11/12pii)#, #[mod 2pi]#

Explanation:

The exponential form of a complex number #z=a+ib# is

#z=re^(itheta)#

where,

#r=|z|=sqrt(a^2+b^2)#

#costheta=a/|z|#

#sintheta=b/|z|#

The complex number is

#z=z_1/z_2=(6-6i)/(-sqrt3+i)#

For the numerator,

#z_1=6-6i=6sqrt2(1/sqrt2-1/sqrt2i)#

#costheta_1=1/sqrt2#

#sintheta_1=-1/sqrt2#

#theta_1=-1/4pi# #[mod 2pi]#

#z_1=6sqrt2e^(-pi/4i)#

For the denominator,

#z_2=-sqrt3+i=2(-sqrt3/2+1/2i)#

#costheta_2=-sqrt3/2#

#sintheta_2=1/2#

#theta_2=5/6pi#, #[mod 2pi]#

#z_=2e^(5/6pii)#

Finally,

#z=z_1/z_2=(6sqrt2e^(-pi/4i))/(2e^(5/6pii))#

#=3sqrt2e^((-pi/4-5/6pi)i)#

#=3sqrt2e^(-13/12pii)#

#=3sqrt2e^(11/12pii)#, #[mod 2pi]#