Question

How do you write `-i` in trigonometric form?

Answer 1

`\qquad \qquad \qquad \qquad \ \ -i \ = \ cos \pi + i sin \pi.`

Explanation:

`"One way to do this is as follows."`

`"If" \ \ z \ \ "is a complex number, then:"`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad z \ = \ r ( cos theta + i sin theta ); \qquad \quad "where:"`

`\quad r = "magnitude (length) of" \ \ z,`

`\quad \theta = "the angle" \ \ z \ \ "makes, in radians when" \ z \ "is placed in"`
`\qquad \qquad \qquad "standard position in the complex plane."`

`"Now visualize" \ z = -i \ \ "on the complex plane. It lies on the"`
`"negative" \ x "-axis, 1 unit in length, to the left of the origin. The"`
`"angle of" \ -i \ \ "in this, its standard position, is clearly" \ \ \pi \ \"radians."`

`"So now we see, for" \ z = -i :`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad r = 1,`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \theta = \pi.`

`"Thus:"`

`\qquad \qquad \qquad \qquad \qquad \qquad \quad -i \ = \ 1 cdot ( cos \pi + i sin \pi )`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ \ = \ cos \pi + i sin \pi.`

`"So:"`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad -i \ = \ cos \pi + i sin \pi.`

`"This is our answer."`