Question

find the cube root of complex number z = -64

Answer 1

Answer:

Step-by-step explanation:

The cube root of a complex number z can be found by expressing z in polar form and then using the cube root property. The polar form of a complex number z = a + bi is given by:

$z = r(\cos(\theta) + i\sin(\theta))$

where r is the magnitude of the complex number $r = \sqrt{a^2 + b^2}$and $theta$is the argument of the complex number $theta = \arctan\left(\frac{b}{a}\right$).

For the complex number z = -64, we have a = -64 and b = 0. Therefore, $r = \sqrt{(-64)^2 + 0^2} = 64$ and $theta = \arctan\left(\frac{0}{-64}\right) = -\frac{\pi}{2}$.

So, the polar form of z is:

$z = 64(\cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2}))$

Now, the cube root property for complex numbers states that the cube roots of a complex number w can be found by taking the cube roots of its magnitude and dividing its argument by 3. Therefore, the cube root of z is:

$sqrt[3]{z} = \sqrt[3]{64}(\cos\left(\frac{-\frac{\pi}{2}}{3}\right) + i\sin\left(\frac{-\frac{\pi}{2}}{3}\right))$ \]

Now, simplify this expression to get the cube root of the complex number z = -64.