Find three cubed roots for the complex number:
125/2+125/2 i√3.
answer with r=
and theta = ___degrees ___degrees and ___degrees
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.
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°
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.
To find the cube roots of a complex number, we can use the polar form of the number. The polar form of a complex number is given by:
z = r(cos(theta) + isin(theta))
where r is the magnitude of the complex number, and theta is the argument or angle.
To convert the given complex number into polar form, we need to find r and theta. Let's break down the given complex number:
125/2 + (125/2) * i√3
To find the magnitude (r), we use the formula:
r = √(real part)^2 + (imaginary part)^2
In this case, the real part is 125/2, and the imaginary part is (125/2) * √3. Substituting these values, we get:
r = √((125/2)^2 + ((125/2) * √3)^2)
Simplifying this expression, we find:
r = √(15625/4 + 9375/4 * 3)
r = √(15625/4 + 28125/4)
r = √43750/4
r = √43750 / 2
r = 125 / √2
Now, to find theta, we use the formula:
θ = arctan((imaginary part) / (real part))
In this case, the imaginary part is (125/2) * √3, and the real part is 125/2. Substituting these values, we get:
θ = arctan(((125/2) * √3) / (125/2))
Simplifying this expression, we find:
θ = arctan(√3)
To find the three cube roots of the complex number, we can use the polar form:
For the principal cube root, we keep r the same and divide theta by 3:
r = 125 / √2
theta = (arctan(√3) / 3)
For the second cube root, we add 120 degrees to the original theta:
r = 125 / √2
theta = (arctan(√3) + 120)
For the third cube root, we add 240 degrees to the original theta:
r = 125 / √2
theta = (arctan(√3) + 240)
So, the three cube roots in polar form are:
1) r = 125 / √2, theta = (arctan(√3) / 3)
2) r = 125 / √2, theta = (arctan(√3) + 120)
3) r = 125 / √2, theta = (arctan(√3) + 240)
Note: The angles are given in radians.