Question

How do you express the complex number in trigonometric form: -6i?

Answer 1

`-6i = 6(cos((3pi)/2) + isin((3pi)/2))`

Explanation:

Trigonometric form is where

`a + bi = absz(cosx + isinx)`.

`absz` is given by Pythagoras', `sqrt(a^2+b^2)`, using the values from `a + bi`.

`a + bi = 0 - 6i`
`sqrt(a^2+b^2) = sqrt(0^2 + (-6)^2) = sqrt36 = 6`

Now you can divide both sides by `6`, so

`(-6i)/6 = (6(cosx + isinx))/6`
`-i = cosx + isinx`

This gives us, obviously, that

`cosx = 0`
`sinx = -1`

and so, superimposing the graphs and finding points that satisfy the simultaneous equations,

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At the point `x = 270`, the `cosx` graph is `0` and the `sinx` graph is at `-1`, which satisfies the equations. Being repeating patterns, obviously there will be other results, such as `x = -450, -90, 630, 990` etc.

Putting all of this back together,

`-6i = 6(cos270 + isin270)`

or

`-6i = 6(cos((3pi)/2) + isin((3pi)/2))`