How do you divide # (1-3i) / (-2+i) # in trigonometric form?

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Answer 1 · 2018-02-24T16:29:16

#(1-3i)/(-2+i)=sqrt2(cos((3pi)/4)+isin((3pi)/4))#

For the complex numbers #z# and #w#,

#|z/w|=|z|/|w|#

and

#arg(z/w)=arg(z)-arg(w)#

So let #z=1-3i# and let #w=-2+i#

#|z|=sqrt(1^2+3^2)#
#=sqrt10#

enter image source here

#tantheta=3/1#
#theta=tan^-1(3)#
#:.arg(z)=-tan^-1(3)#

#|w|=sqrt(2^2+1^2)#
#=sqrt5#

enter image source here

#tantheta=1/2#
#theta=tan^-1 (1/2)#
#:. arg(w)=pi-tan^-1 (1/2)#

#:. |z/w|=sqrt10/sqrt5#
#=sqrt(10/5)#
#=sqrt2#

#arg(z/w)=-tan^-1(3)-(pi-tan^-1 (1/2))#
#=tan^-1(1/2)-tan^-1 (3)-pi#
#=-5/4pi#

since #-pi < arg(z/w) <= pi#

#arg(z/w)=(3pi)/4#

#:. z/w=sqrt2(cos((3pi)/4)+isin((3pi)/4))#