How do you divide # (3+4i) / (1+4i) # in trigonometric form?

1 Answer
Jul 28, 2017

The answer is #=5/sqrt17(0.92-0.39i)#

Explanation:

We start by writing numerator and denominator in polar form

#z=z_1/z_2#

The polar form of a complex number is

#z=r(costheta+isintheta)....................#(1)#

The numerator is

#z_1=3+4i#

#r_1=|z_1|=sqrt((3)^2+(4)^2)=sqrt(9+16)=sqrt25=5#

Therefore,

#z_1=5(3/5+4/5i)#

Comparing this equation to equation #(1)#

#costheta=3/5# and #sintheta=4/5#

So,

we are in the quadrant #I#

#theta=53.13^@#

The polar form is

#z_1=5(cos(53.13^@)+isin(53.13^@))=5e^(53.13i)#

The denominator is

#z_2=1+4i#

#r_2=|z_2|=sqrt((1)^2+(4)^2)=sqrt(1+16)=sqrt17#

Therefore,

#z_2=sqrt17(1/sqrt17+4/sqrt17i)#

Comparing this equation to equation #(1)#

#costheta=1/sqrt17# and #sintheta=4/sqrt17#

So,

we are in the quadrant #I#

#theta=75.96^@#

The polar form is

#z_2=sqrt17(cos(75.96^@)+isin(75.96^@))=sqrt17e^(75.96i)#

Terefore,

#z=z_1/z_2=(5e^(53.13i))/(sqrt17e^(75.96i))#

#=(5/sqrt17)e^((53.13-75.96)i)#

#=(5/sqrt17)e^((-22.83^@)i)#

#=5/sqrt17(cos(-22.83^@)+isin(-22.83^@))#

#=5/sqrt17(0.92-0.39i)#