What is the trigonometric form of # (4-2i) #?

2 Answers
Fleur
Jul 13, 2017

#2sqrt5# #cis# #(-0.46)#

Explanation:

To find the trigonometric form, we have to know #r#, the distance of the point from the origin, and #theta#, the angle.

We can use the following formulas:

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

#tan theta = b/a#

#r = sqrt(4^2+(-2)^2)#
#r = sqrt(20)#
#r = 2sqrt(5)#

#tan theta = -2/4#
#theta = tan^-1(-2/4)#
#theta = -0.46#

So, the trigonometric form is #2sqrt5# #cis# #(-0.46)# or #2sqrt5 (cos (-0.46) + i sin (-0.46))#.

Jim G.
Jul 13, 2017

#2sqrt5(cos(0.46)-isin(0.46))#

Explanation:

#"to convert from "color(blue)"cartesian to trig. form"#

#"that is " (x,y)tor(costheta+isintheta)" use"#

#•color(white)(x)r=sqrt(x^2+y^2)#

#•color(white)(x)theta=tan^-1(y/x)color(white)(x);-pi < theta<=pi#

#"here "x=4" and " y=-2#

#rArrr=sqrt(4^2+(-2)^2)=sqrt20=2sqrt5#

4-2i is in the fourth quadrant so we must ensure that #theta# is in the fourth quadrant.

#theta=tan^-1(1/2)=0.46larrcolor(red)" related acute angle"#

#rArrtheta=-0.46larrcolor(red)" in fourth quadrant"#

#rArr4-2i=2sqrt5(cos(-0.46)+isin(-0.46))#

#rArr4-2i=2sqrt5(cos(0.46)-isin(0.46))#

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