Question

How do you write the trigonometric form of `1+6i`?

Answer 1

`sqrt(37)(cos 80.5^o + i*sin 80.5^o) or sqrt(37)*cis 80.5^o`

Explanation:

Converting complex numbers into trigonometric form is very similar to converting rectangular coordinates (aka Cartesian Coordinates) into polar coordinates. There are 2 formulas that will assist (assuming the complex number is in `a+bi` format):

`r = sqrt(a^2+b^2)`

`tan theta_{ref} = |b/a|`

We begin by finding the value `r`:

`r = sqrt(1^2 + 6^2) = sqrt(1+36) = sqrt(37)`

Next, we can determine the reference angle. For this problem, I am using degrees, though the same work could be done with radians as the measure instead.

`tan theta_{ref} = |6/1|`

`theta_{ref} = 80.5^o`

This is only the reference angle, which will always be in the first quadrant when done this way. The complex number `1+6i` lies in the first quadrant in the complex plane, and so this angle will be sufficient for our needs.

The trigonometric form of a complex number `a + bi` is written one of two ways:

`{(r(cos theta + i*sin theta), "Expanded Form"),(r*cis theta, "Condensed Form"):}`

Thus, our answer is:

`sqrt(37)(cos 80.5^o + i*sin 80.5^o) or sqrt(37)*cis 80.5^o`