Question

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

Answer 1

`z= sqrt26/2(cos(78.69^@)+isin(78.69^@))` or
`z=5/2i+1/2`

Explanation:

Let's start by solving this in the form given and see what we get:

Multiply the conjugate of the denominator:
`( -3i+2) / (-1-i)*(-1+i)/(-1+i)=(3i-2-3i^2+2i)/(1-i^2)=(5i+1)/2=5/2i+1/2`

`1/2+5/2i`
We can convert this to trigonometric form by finding `r`:
`r=sqrt(a^2+b^2)`
`theta=arctan(b/a)`

`r=sqrt((1/2)^2+(5/2)^2)`

`r= sqrt26/2`

`theta=arctan(5)`
`thetaapprox78.69^@`

So we can say the solution in trig form is
`z=sqrt26/2(cos(78.69^@)+isin(78.69^@))`

Let's double check by solving in trigonometric form:
`z_1= 2-3i`
`r= sqrt((2)^2+(-3)^2)`
`r= sqrt(13)`

`theta= arctan(-3/2)approx-56.31^@approx303.69^@`

`z_2=-1-i`
`r= sqrt((-1)^2+(-1)^2)`
`r=sqrt2`
`theta= arctan(-1/-1)= 45^@`, however this in the third quadrant so`+180^@= 225^@`

`z_1/z_2= (sqrt13(cos(303.69^@)+isin(303.69^@)))/(sqrt2(cos(225^@)+isin(225^@))`

WHEN DIVIDING IN TRIGONOMETRIC FORM:
Divide the r by the other r
And subtract the angles

`sqrt13/sqrt2*sqrt2/sqrt2= sqrt26/2`

`303.69^@-225^@= 78.69^@`

So same solution:
`z= sqrt26/2(cos(78.69^@)+isin(78.69^@))`