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

1 Answer
Feb 24, 2018

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

Explanation:

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

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

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

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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))