Let z1=2(cos 2pi/3+isin 2pi/3)
what are z1^(1/3)?
To find z1^(1/3), we first need to express z1 in exponential form using Euler's formula:
Euler's formula states that for any complex number z = x + yi,
z = r(cosθ + isinθ),
where r = √(x^2 + y^2) is the modulus (magnitude) of z, and θ = atan2(y, x) is the argument (angle) of z.
In this case, z1 = 2(cos(2π/3) + isin(2π/3)).
We can convert this to exponential form as follows:
z1 = 2 * exp(i * (2π/3)),
where exp denotes the exponential function.
Now, let's find the cube root of z1.
To find the cube root of z1, we need to find the cube root of its magnitude and divide the argument by 3.
The magnitude of z1 is:
|z1| = |2 * exp(i * (2π/3))| = 2.
The argument of z1 is:
arg(z1) = arg(2 * exp(i * (2π/3))) = 2π/3.
Now, let's find the cube root of the magnitude:
|z1^(1/3)| = ∛|z1| = ∛2.
And, let's find the argument of the cube root:
arg(z1^(1/3)) = arg(z1) / 3 = (2π/3) / 3 = 2π/9.
Finally, we can express z1^(1/3) in exponential form:
z1^(1/3) = |z1^(1/3)| * exp(i * arg(z1^(1/3))) = (∛2) * exp(i * (2π/9)).
Therefore, z1^(1/3) = (∛2) * exp(i * (2π/9)).