plz find 3 cube roots of -2+2√3 i
put this into polar form
r = root( -2^2 + (2root(3))^2)
= root( 4 + 6)
= root(10)
tan theta = 2root(3)/-2
theta = arctan (-root(3))
theta = pi/3
polar form is
root(10) cis (pi/3)
this is the first root
the second root is
root(10) cis (2pi + pi/3)
the third root is
root(10) cis (4pi + pi/3)
Let z = -2+2√3 i
Converting to polar form:
r^2 = (-2)^2 + (2√3)^2 = 4 + 4*3 = 16
r = 4
arg = 2pi/3
z^(1/3) = r^(1/3)e^(i*arg/3) = 4^(1/3) e^(2pi/9*i) = 2^(2/3)[cos(2pi/9) + i*sin(2pi/9)]
The other two are r^(1/3)e^(i*(2pi +arg)/3) = 2^(2/3)[cos(8pi/9) + i*sin(8pi/9)]
and r^(1/3)e^(i*(4pi +arg)/3) = 2^(2/3)[cos(14pi/9) + i*sin(14pi/9)]
Find cube roots of -2 + 2√3 i.
(-2 + 2√3.i)^(1/3) = [sqrt((-2)^2 +2√3)^2).e^i(-arctan√3)]^(1/3)
(-2 + 2√3.i)^(1/3) = [4.e^(i.2pi/3)]^(1/3)
(-2 + 2√3.i)^(1/3) = (4^1/3).e^i.2pi/9 = 1.587e^i.2pi/9
(-2 + 2√3.i)^(1/3) = 1.587(0.766 + i.0.643)
(-2 + 2√3)^(1/3) = 1.216 + i.1.020 >===============< ANSWER
Yes the answer is x -5
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put this into polar form
r = root( -2^2 + (2root(3))^2)
= root( 4 + 6)
= root(10)
tan theta = 2root(3)/-2
theta = arctan (-root(3))
theta = pi/3
polar form is
root(10) cis (pi/3)
this is the first root
the second root is
root(10) cis (2pi + pi/3)
the third root is
root(10) cis (4pi + pi/3)
Let z = -2+2√3 i
Converting to polar form:
r^2 = (-2)^2 + (2√3)^2 = 4 + 4*3 = 16
r = 4
arg = 2pi/3
z^(1/3) = r^(1/3)e^(i*arg/3) = 4^(1/3) e^(2pi/9*i) = 2^(2/3)[cos(2pi/9) + i*sin(2pi/9)]
The other two are r^(1/3)e^(i*(2pi +arg)/3) = 2^(2/3)[cos(8pi/9) + i*sin(8pi/9)]
and r^(1/3)e^(i*(4pi +arg)/3) = 2^(2/3)[cos(14pi/9) + i*sin(14pi/9)]
Find cube roots of -2 + 2√3 i.
(-2 + 2√3.i)^(1/3) = [sqrt((-2)^2 +2√3)^2).e^i(-arctan√3)]^(1/3)
(-2 + 2√3.i)^(1/3) = [4.e^(i.2pi/3)]^(1/3)
(-2 + 2√3.i)^(1/3) = (4^1/3).e^i.2pi/9 = 1.587e^i.2pi/9
(-2 + 2√3.i)^(1/3) = 1.587(0.766 + i.0.643)
(-2 + 2√3)^(1/3) = 1.216 + i.1.020 >===============< ANSWER
Yes the answer is x -5