Question

Find three cubed roots for the complex number:
125/2+125/2 i√3.
answer with r=
and theta = ___degrees ___degrees and ___degrees

Answer 1

AAAaannndd the bot gets it wrong yet again!

125/2 + 125/2i√3 = 125 cis π/3
so the cube roots are
5 cis π/9 + k* 2π/3, k=0,1,2
or
5 cis 20°, 140°, 260°

Answer 2

One way to find the cubed roots of a complex number is to use the polar form:

z = r(cos theta + i sin theta)

where r is the modulus or absolute value of z, and theta is the argument or angle that z makes with the positive real axis.

To find the cube roots, we can take the cube root of r and divide theta by 3. Then we can find the three cube roots by adding 2pi/3 and 4pi/3 to the divided angle.

In this case, we have:

z = 125/2 + 125/2i√3

r = sqrt((125/2)^2 + (125/2√3)^2) = 125/2 * sqrt(2)

theta = arctan((125/2√3)/(125/2)) = pi/3

So, using the formula for the cube roots, we get:

r1 = (125/2 * sqrt(2))^(1/3) = 5 * (2^(2/3))

theta1 = pi/9 radians = 20 degrees

z1 = r1(cos theta1 + i sin theta1)

r2 = (125/2 * sqrt(2))^(1/3) = 5 * (2^(2/3))

theta2 = (pi/9 + 2pi/3) radians = 100 degrees

z2 = r2(cos theta2 + i sin theta2)

r3 = (125/2 * sqrt(2))^(1/3) = 5 * (2^(2/3))

theta3 = (pi/9 + 4pi/3) radians = 220 degrees

z3 = r3(cos theta3 + i sin theta3)

Therefore, the three cubed roots of the complex number 125/2+125/2i√3 are:

5(2^(2/3)) cis 20 degrees, 5(2^(2/3)) cis 100 degrees, and 5(2^(2/3)) cis 220 degrees.

Answer 3

I apologize for the mistake in my previous response. You are correct that the cube roots of 125/2 + 125/2i√3 are 5 cis 20°, 140°, and 260°. Thank you for pointing out the error.