Find all the roots of :z^3=2-3i ,and leave your answer in rectangular form ?

To find the roots of the equation z^3 = 2 - 3i, we first need to convert the complex number 2 - 3i to polar form.

Let z = r(cosθ + isinθ), where r is the magnitude of z and θ is the argument of z.

Given z^3 = 2 - 3i, we have:

r^3(cos(3θ) + isin(3θ)) = 2 - 3i

Now, equate the real and imaginary parts on both sides:

r^3cos(3θ) = 2
r^3sin(3θ) = -3

From the first equation, we get:

r^3 = 2
r = ∛2

From the second equation, we get:

sin(3θ) = -3/∛2

Now, we can find the values of θ by solving for θ:

3θ = sin^(-1)(-3/∛2) + 2πk
θ = (sin^(-1)(-3/∛2) + 2πk)/3

Now, substitute the values of r and θ back into z = r(cosθ + isinθ) to find the roots in rectangular form.

$$z^3=2-3i$$

$$z^3=2\sqrt{2}\left ( \cos \frac{-\pi}{4}+i\sin \frac{-\pi}{4} \right )$$

So the three cube roots of \(2\sqrt{2}\left ( \cos \frac{-\pi}{4}+i\sin \frac{-\pi}{4} \right )\) are:

$$z_1=\sqrt[3]{2}\left ( \cos \left ( \frac{-\pi}{12} \right )+i\sin \left ( \frac{-\pi}{12} \right ) \right )$$

$$z_2=\sqrt[3]{2}\left ( \cos \left ( \frac{-\pi}{12}-\frac{2\pi}{3} \right )+i\sin \left ( \frac{-\pi}{12}-\frac{2\pi}{3} \right ) \right )$$

$$z_3=\sqrt[3]{2}\left ( \cos \left ( \frac{-\pi}{12}+\frac{2\pi}{3} \right )+i\sin \left ( \frac{-\pi}{12}+\frac{2\pi}{3} \right ) \right )$$