cube roots of 64i?
Well, as a mathematically inclined clown, I must clarify that the cube root of a complex number can have multiple answers. In this case, the cube roots of 64i are not as tasty as ice cream, but they are quite interesting. Let's break it down, shall we?
First, we can write 64i in its polar form. The magnitude (r) of 64i is 64, and the argument (θ) is π/2 (or 90 degrees).
Now, to find the cube root, we divide the magnitude by the cube root of 64, which is 4. Since there are three cube roots, we can multiply the argument by 1/3 and add multiples of 2π/3.
So, the three cube roots of 64i are:
- 4∠(π/6) (or 30 degrees)
- 4∠(7π/6) (or 210 degrees)
- 4∠(11π/6) (or 330 degrees)
Now, go forth and use those cube roots responsibly! Or at least impress your friends at the next math party.
To find the cube roots of 64i, we can start by converting 64i to polar form.
Let's first find the modulus, which is the distance from the origin in the complex plane. The modulus can be calculated using the formula: |z| = √(Re^2 + Im^2), where Re is the real part and Im is the imaginary part.
For 64i, Re = 0 and Im = 64, so |64i| = √(0^2 + 64^2) = 64.
Now, let's find the argument, which represents the angle with respect to the positive real axis. The argument can be calculated using the formula: θ = arctan(Im/Re).
For 64i, Re = 0 and Im = 64, so θ = arctan(64/0) = π/2.
Now we have 64i in polar form: |64i|∠θ = 64∠(π/2).
To find the cube roots, we can use De Moivre's theorem, which states that for any complex number z in polar form (r∠θ), the nth roots can be found using the formula: z^(1/n) = r^(1/n)∠(θ/n).
Using this formula, we can find the cube roots of 64i:
1. First cube root: (64∠(π/2))^(1/3) = 4∠(π/6)
2. Second cube root: (64∠(π/2))^(1/3) = 4∠(π/2 + 2π/3) = 4∠(5π/6)
3. Third cube root: (64∠(π/2))^(1/3) = 4∠(π/2 + 4π/3) = 4∠(7π/6)
So, the three cube roots of 64i are: 4∠(π/6), 4∠(5π/6), and 4∠(7π/6).
To find the cube roots of 64i, we can start by expressing 64i in its polar form. In polar form, a complex number is represented as z = r(cosθ + isinθ), where r is the magnitude of the number and θ is the angle it makes with the positive real axis.
To find the magnitude (r) and argument (θ) of 64i, we can use the following formulas:
r = √(a^2 + b^2) (The magnitude of a complex number z = a + bi is given by r = √(a^2 + b^2). In this case, a = 0 and b = 64.)
θ = arctan(b/a) (The argument of a complex number z = a + bi is given by θ = arctan(b/a). In this case, a = 0 and b = 64.)
So, let's calculate the values:
r = √(0^2 + 64^2) = 64
θ = arctan(64/0) = π/2 or 90 degrees
Now that we have the polar form of 64i, which is 64(cos(π/2) + isin(π/2)), we can find the cube roots by using the formula:
z^3 = r^3(cos(3θ) + isin(3θ))
Substituting the values, we get:
z^3 = 64^3(cos(3(π/2)) + isin(3(π/2)))
Simplifying further:
z^3 = 262144(cos(3π/2) + isin(3π/2))
Now, let's find the values of the cube roots:
z1 = ∛(262144)(cos((3π/2)/3) + isin((3π/2)/3))
z2 = ∛(262144)(cos((3π/2 + 2π)/3) + isin((3π/2 + 2π)/3))
z3 = ∛(262144)(cos((3π/2 + 4π)/3) + isin((3π/2 + 4π)/3))
Calculating each of the cube roots separately using a calculator or by hand will give you the three cube roots of 64i.
β=90° (sin-164/64=sin-11=π/2)
Convert the given number into the form
a+bi = α(cos(β)+i sin(β))
where
α = sqrt(a^2+b^2)
β = sin-1(b/α)
If the number is plot in the complex (Z) plane, it will be more evident.
For example, 64i will have
α=64 (the distance from origin)
β=90° (sin-164/64=1)
The equivalent angles are:
90°
450°
810°
(we only need 3)
Divide each of these angles by , to get β1, β2, & β3. And take α'=α^(1/3)=64^(1/3)=4
The three cube roots are then
z1=α'(cos(β1)+isin(β1)
=4(cos(30°)+isin(30°))
z2=4(cos(150°)+isin(150°))
z3=4(cos(270°)+isin(270°))
Check by expanding z1^3, z2^3 and z3^3 to get back 64i.