`1 - sqrt(3)i` Write the trigonometric form of the number.

Given: `1-sqrt(3)i`

`r=sqrt(1^2+(-sqrt(3))^2)=sqrt(1+3)=sqrt(4)=2`

`tantheta=-sqrt3/1=-sqrt3`

Since the angle is in quadrant 4,`<br> `

`theta=arctan(-sqrt(3))=(5pi)/3`

In trigonometric form `z=2[cos((5pi)/3)+isin((5pi)/3)]`

Given: `1-sqrt(3)i`

`r=sqrt(1^2+(-sqrt(3))^2)=sqrt(1+3)=sqrt(4)=2`

`tantheta=-sqrt3/1=-sqrt3`

Since the angle is in quadrant 4,`<br> `

`theta=arctan(-sqrt(3))=(5pi)/3`

In trigonometric form `z=2[cos((5pi)/3)+isin((5pi)/3)]`