Find all cube roots of the complex number 64(cos(219°)+isin(219°)) . Leave answers in polar form and show all work.
To find all cube roots of the complex number, we first need to express the complex number in polar form.
Let's solve for the magnitude and argument of the complex number:
Magnitude: |z| = √(64^2) = 64
Argument: arg(z) = 219°
Now, the complex number can be expressed as:
64(cos(219°) + isin(219°)) = 64cis(219°)
Next, let's find the cube roots of the complex number by using De Moivre's theorem:
Let z = 64cis(219°)
Then, the cube roots can be found as:
√z = √(64cis(219°)) = 64^(1/3)cis(219°/3)
= 4cis(73°)
Therefore, the cube roots of the complex number 64(cos(219°) + isin(219°)) are:
4cis(73°) + 4cis(73° + 120°) + 4cis(73° + 240°)
This simplifies to:
4cis(73°) + 4cis(193°) + 4cis(313°)
Therefore, the cube roots of the complex number in polar form are:
4cis(73°), 4cis(193°), and 4cis(313°).
can you explain in simpler terms
Sure! When finding the cube roots of a complex number, we first need to convert the complex number from rectangular form to polar form. In this case, the complex number is 64(cos(219°) + isin(219°)), which can be written in polar form as 64cis(219°).
To find the cube roots of this complex number, we apply De Moivre's theorem. We take the cube root of the magnitude (64) and divide the argument (219°) by 3. This gives us the magnitude and argument of the cube roots.
The cube roots of the complex number in polar form are then computed as:
- Cube root 1: 4cis(73°)
- Cube root 2: 4cis(193°)
- Cube root 3: 4cis(313°)
These are the three cube roots of the given complex number in polar form.
What is cis
"Cis" is a shorthand way to express complex numbers in polar form. It stands for "cosine + i sine" and is commonly used in mathematics.
For example, a complex number in polar form, such as z = r(cosθ + i sinθ), can be written as z = rcisθ.
So, when we say 64cis(219°), we are referring to the complex number 64(cos(219°) + isin(219°)) written in a shorthand form using "cis."