What is the trigonometric form of (2-12i) (212i)?

1 Answer
Jan 31, 2016

2sqrt(37)(cos(arctan(-6))+isin(arctan(-6))237(cos(arctan(6))+isin(arctan(6))
color(white)("XXXXXXXXXX")~~12.166(0.1644-i*0.9864)XXXXXXXXXX12.166(0.1644i0.9864)

Explanation:

For a complex number in the form: a+bia+bi
its trigonometric form is:
color(white)("XXX")|z|(cos(theta)+i*sin(theta))XXX|z|(cos(θ)+isin(θ))
where
color(white)("XXX")|z| = sqrt(a^2+b^2)XXX|z|=a2+b2
and
color(white)("XXX")theta = {(arctan(b/a),"if " a+bi " in QI or QIV"),(pi+arctan(b/a), "if "a+bi " in QII or QIII"):}

Given (2-12i)
color(white)("XXX")|z| =sqrt(2^2+(-12)^2) =4sqrt(37)

and since (2-12i) is in QII
color(white)("XXX")theta = arctan(-6)