find the 2 square roots of -36+36(sqrt)3i
We can begin by factoring out 36 from the expression to get:
-36 + 36√3i = 36(-1 + √3i)
Now we can take the square root of both sides:
√(-36+36√3i) = √(36(-1 + √3i))
The square root of 36 is 6, so we can simplify further:
√(-36+36√3i) = 6√(-1 + √3i)
To find the two square roots of the expression, we will multiply 6 by the square roots of -1+√3i. Let's call these roots x and y. We know that x * y = -1+√3i and x^2 = y^2 = -1+√3i. We can solve for x and y using the quadratic formula:
x = (1 + √3i ± √(-3 + 2√3i)) / 2
y = (1 + √3i ∓ √(-3 + 2√3i)) / 2
Simplifying the complex numbers under the square root, we get:
x = (1 + √3i ± i√(3 + 2√3i)) / 2
y = (1 + √3i ∓ i√(3 + 2√3i)) / 2
Therefore, the two square roots of -36+36√3i are:
6[(1 + √3i + i√(3 + 2√3i)) / 2]
and
6[(1 + √3i - i√(3 + 2√3i)) / 2]
find the three cube roots of -1 write the answer in polar form.
The cube roots of -1 are the solutions to the equation z^3 = -1. We can write -1 in polar form as 1 * e^(iπ), so we are looking for complex numbers z that satisfy the equation z^3 = 1 * e^(iπ).
To find these cube roots, we can write z in polar form as re^(iθ), where r is the magnitude of z and θ is the angle that z makes with the positive real axis. Then we can use the fact that (re^(iθ))^3 = r^3 * e^(3iθ) and solve for r and θ.
Since we want z^3 to equal -1, we also know that r^3 = 1 and 3θ = π + 2nπ, where n is an integer. Solving for r and θ gives us:
r = 1, θ = (π + 2nπ) / 3
There are three distinct values of θ that satisfy this equation for n = 0, 1, and 2:
θ1 = π / 3
θ2 = π + π / 3 = 4π / 3
θ3 = 2π + π / 3 = 7π / 3
Plugging these values of θ into the polar form of z, we get the three cube roots of -1:
z1 = 1 * e^(iπ/3) = cos(π/3) + i*sin(π/3) = (1/2 + √3/2 i)
z2 = 1 * e^(i4π/3) = cos(4π/3) + i*sin(4π/3) = (-1/2 + √3/2 i)
z3 = 1 * e^(i7π/3) = cos(7π/3) + i*sin(7π/3) = (-1 + 0i)
Therefore, the three cube roots of -1 are (1/2 + √3/2 i), (-1/2 + √3/2 i), and (-1 + 0i), which we have written in polar form.
find the three cube roots of -1 write the answer in polar form with degrees
To express the answer in degrees, we need to convert the angles from radians to degrees. We know that π radians is equivalent to 180 degrees, so we can use the conversion factor 180/π to convert the angles.
The cube roots of -1 are the solutions to the equation z^3 = -1. We can write -1 in polar form as 1 * e^(i180), so we are looking for complex numbers z that satisfy the equation z^3 = 1 * e^(i180).
To find these cube roots, we can write z in polar form as re^(iθ), where r is the magnitude of z and θ is the angle that z makes with the positive real axis. Then we can use the fact that (re^(iθ))^3 = r^3 * e^(3iθ) and solve for r and θ.
Since we want z^3 to equal -1, we also know that r^3 = 1 and 3θ = 180 + 360n, where n is an integer. Solving for r and θ gives us:
r = 1, θ = (180 + 360n) / 3
There are three distinct values of θ that satisfy this equation for n = 0, 1, and 2:
θ1 = 60 degrees
θ2 = 240 degrees
θ3 = 420 degrees
Plugging these values of θ into the polar form of z, we get the three cube roots of -1:
z1 = 1 * e^(i60) = cos(60) + i*sin(60) = (1/2 + √3/2 i)
z2 = 1 * e^(i240) = cos(240) + i*sin(240) = (-1/2 + -√3/2 i)
z3 = 1 * e^(i420) = cos(420) + i*sin(420) = (-1 + 0i)
Therefore, the three cube roots of -1 are (1/2 + √3/2 i), (-1/2 - √3/2 i), and (-1 + 0i), expressed in polar form with degrees.
AAAaannndd the bot gets it wrong yet again!
As for √(-36+36√3i)
your answer is wrong, since
(6[(1 + √3i - i√(3 + 2√3i)) / 2])^2 = 55.210-57.411i
What you should have done was
36(-1 + √3i) = 36√2 cis(2π/3)
so
√(-36+36√3i) = 6∜2 cis(π/3) = 6∜2 (±1+√3 i)/2
You are correct! I apologize for the mistake in my previous answer. Thank you for catching it and providing the correct solution.
To find the square roots of -36 + 36√3i, we can begin by factoring out 36 to get:
-36 + 36√3i = 36(-1 + √3i)
We can then write this expression in polar form as:
36√2 cis(2π/3)
To find the square roots, we need to take the square root of both the magnitude and the angle. The magnitude of the square root is the square root of the magnitude of the original number, which is 36√2. The angle of the square root is half the angle of the original number, which is π/3.
Therefore, the two square roots are:
6√2 cis(π/6) = 6√2 (cos(π/6) + i sin(π/6)) = 3√2 + 3i
-6√2 cis(π/6) = -6√2 (cos(π/6) + i sin(π/6)) = -3√2 - 3i
Thank you again for pointing out my error.