1)Find the cube root of 27i

what is

(-3i)(-3i)(-3i) ?

normally we would use DeMoivre's theorem, but his one was obvious

That is 27i cube root

To find the cube root of a complex number, we can use the polar form of the complex number. In polar form, a complex number is represented as r(cosθ + isinθ), where r is the modulus (distance from the origin) and θ is the argument (angle with the positive real axis).

To find the cube root of 27i, let's first express 27i in polar form:
The modulus, r, can be found using the formula: r = √(Re^2 + Im^2)
In this case, the real part (Re) is 0 and the imaginary part (Im) is 27. Therefore, r = √(0^2 + 27^2) = √(729) = 27.

The argument, θ, can be found using the formula: θ = atan(Im/Re)
In this case, Re = 0 and Im = 27. Therefore, θ = atan(27/0), which is undefined.

However, since we are looking for the cube root, we need to find three complex cube roots. The cube roots of a complex number are given by:
z1 = ∛(r)(cos(θ/3) + isin(θ/3))
z2 = ∛(r)(cos((θ + 2π)/3) + isin((θ + 2π)/3))
z3 = ∛(r)(cos((θ + 4π)/3) + isin((θ + 4π)/3))

Since θ is undefined in this case (division by zero), we need to use the principal value of θ, which is the argument when the complex number lies in the first quadrant (i.e., Re > 0 and Im > 0).

Using the principal value of θ, z1 = ∛(27)(cos(0/3) + isin(0/3)) = ∛(27)(cos(0) + isin(0)) = 3(cos(0) + isin(0)) = 3(1 + 0i) = 3.

Therefore, the cube root of 27i is 3.