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

1 Answer
Alan P.
Jan 31, 2016

#2sqrt(37)(cos(arctan(-6))+isin(arctan(-6))#
#color(white)("XXXXXXXXXX")~~12.166(0.1644-i*0.9864)#

Explanation:

For a complex number in the form: #a+bi#
its trigonometric form is:
#color(white)("XXX")|z|(cos(theta)+i*sin(theta))#
where
#color(white)("XXX")|z| = sqrt(a^2+b^2)#
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)#

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