How do you find the square root of #16 ( cos ((2pi)/3) + i sin ((2pi)/3))#?

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Answer 1 · 2016-07-05T21:16:41

#sqrt(16(cos((2pi)/3)+i sin((2pi)/3))) = 4(cos(pi/3) + i sin(pi/3)) = 2+2sqrt(3)i#

If #r >= 0# and #-pi < theta <= pi# then:

#sqrt(r(cos theta + i sin theta)) = sqrt(r)(cos (theta/2) + i sin (theta/2))#

So in our example:

#sqrt(16(cos((2pi)/3)+i sin((2pi)/3)))#

#= 4(cos(pi/3)+i sin(pi/3))#

#= 4(1/2+sqrt(3)/2i)#

#=2+2sqrt(3)i#

Note that this is the principal square root.

The other square root is:

# -4(cos(pi/3)+i sin(pi/3)) = -2-2sqrt(3)i#