Question

How do you write the following in trigonometric form and perform the operation given `(3+4i)/(1-sqrt3i)`?

Answer 1

`z=3/4-root2(3)+i(3/4root2(3)+1)=5/2(costheta+isintheta)`
with `theta=arctan( (25/16root2(3)+3)/(-39/16))`

Explanation:

let's get rid of complex denominator...
`z=(3+4i)/(1-iroot2(3))(1+iroot2(3))/(1+iroot2(3))=(3+i3root2(3)+i4-4root2(3))/4`
`z=3/4-root2(3)+i(3/4root2(3)+1)` that in trigonometric form is `z=abs(z)(costheta+isintheta)`
`abs(z)=root2((3/4-root2(3))^2+(3/4root2(3)+1)^2)`
`abs(z)=root2(9/16+3-3/2root2(3)+27/16+1+3/2root2(3))=`
`abs(z)=root2((36+64)/16)=10/4=5/2`
the angle is `theta=arctan( (3/4root2(3)+1)/(3/4-root2(3)))`
`theta=arctan( (3/4root2(3)+1)/(3/4-root2(3))*(3/4+root2(3))/(3/4+root2(3)))=`

`theta=arctan( (9/16root2(3)+9/4+3/4+root2(3))/(9/16-3))`
`theta=arctan( (25/16root2(3)+3)/(-39/16))`