How do you write the complex number in trigonometric form #-2#?

1 Answer
Dec 28, 2016

#2(cos(pi)+isin(pi))#

Explanation:

#2=2+i * 0 or color(red)2(color(blue)1+i * color(blue)0)#

The general trigonometric for a complex number is
#color(white)("XXX")color(red)r(cos(color(blue)pi)+i * sin(color(blue)pi))#

So we need a value #color(blue)theta#
such that
#color(white)("XXX")cos(color(blue)theta)=1color(white)("XX")andcolor(white)("XX")sin(color(blue)theta)=color(blue)0#

The simplest solution that meets these requirements is #color(blue)theta =color(blue)pi#

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To be complete, we should possibly note that any
#color(white)("XXX")color(blue)theta = color(blue)pi + k2pi, k in ZZ#
would also meet the stated requirement
and the "complete" answer would be
#color(white)("XXX")2(cos(pi+k2pi)+isin(pi+k2pi))), k in ZZ#

While this is correct, I personally find that it obscures the significant components of the answer.