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

1 Answer
Jason K.
Oct 26, 2017

#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#

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