Question

How do you divide `( 2i -4) / ( 5 i -6 )` in trigonometric form?

Answer 1

`(2i-4)/(5i-6)=sqrt(5/61)"cis(tan^-1(-16/29))`

Explanation:

`"Let" z_1=2i-4, "where",`
`r_1=sqrt(2^2+(-1)^2)=sqrt5`
`theta1=tan^-1((-4)/2)`

`"Let" z_2=5i-6, "where",`
`r_1=sqrt(5^2+(-6)^2)=sqrt61`
`theta1=tan^-1((-6)/5)`

`"Now,"`

`z_1=sqrt5("cis"(tan^-1((-4)/2)))`

`z_2=sqrt61("cis"(tan^-1((-6)/5)))`

`(2i-4)/(5i-6)=z_1/z_2=(sqrt5("cis"(tan^-1((-4)/2))))/(sqrt61("cis"(tan^-1((-6)/5))))`
`=(sqrt5)/(sqrt61)("cis"(tan^-1((-4)/2)))/("cis"(tan^-1((-6)/5)))`

`(sqrt5)/(sqrt61)=sqrt(5/61)`
`"By De-Moivre's Theorem,"`

`("cis"(tan^-1((-4)/2)))/("cis"(tan^-1((-6)/5)))=cis(tan^-1((-4)/2)-tan^-1((-6)/5))`

`tan^-1((-4)/2)-tan^-1((-6)/5)=tan^-1(((-4)/2-(-6)/5)/(1+((-4)/2)((-6)/5)))`

`-4/2-6/5=-2-6/5=(-10-6)/5=-16/5`

`1+((-4)/2)((-6)/5)=1+24/5=(5+24)/5=29/5`

Hence,
`tan^-1(((-4)/2-(-6)/5)/(1+((-4)/2)((-6)/5)))=tan^-1((-16/5)/(29/5))=tan^-1(-16/29)`

Thus,

`(2i-4)/(5i-6)=sqrt(5/61)"cis(tan^-1(-16/29))`