How do you write the following in trigonometric form and perform the operation given #(4i)/(-4+2i)#?

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Answer 1 · 2017-03-23T22:57:45

From Cartesian, #a + bi#, to trig #r(cos(theta)+isin(theta))#:

#r = sqrt(a^2+b^2)#
#theta = {(tan^-1(b/a);a>0,b>0),(pi/2; a=0,b>0),(pi+tan^-1(b/a),a<0),((3pi)/2;a=0,b<0), (2pi+tan^-1(b/a);a>0,b<0):}#

For the numerator, #a = 0, b = 4:

#r = sqrt(0^2+4^2)#

#r = 4#
#theta = (pi)/2#

The trig form of the numerator: #4(cos(pi/2)+isin(pi/2))#

For the denominator, #a = -4, b = 2:

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

#r = sqrt(16+4)#

#r = sqrt(20)#

#r = 2sqrt(5)#

#theta = pi + tan(2/-4)#

#theta ~~ 2.68#

The trig form of the numerator: #2sqrt(5)(cos(2.68)+isin(2.68))#

To divide you subtract the angle of the denominator from the angle of the numerator:

#theta = pi/2-2.68#

#theta ~~ -1.1#

And divide the magnitudes:

#r = 4/(2sqrt(5))#

#r = 2sqrt(5)/5#

The trig form is:

#2sqrt(5)/5(cos(-1.1)+isin(-1.1))#