Find all cube roots of the complex number 64(cos219*+isin219*) Leave answers in Polar form and SHOW ALL WORK.
Ps the * is for the degrees sign
To find all cube roots of the complex number 64(cos219° + isin219°), we first need to convert the complex number into its polar form.
Given complex number: 64(cos219° + isin219°)
To find the magnitude r:
r = √(64^2) = 64
To find the argument θ:
θ = tan^(-1)(sin219° / cos219°) = tan^(-1)(1) = 45°
Therefore, the polar form of the complex number is 64(cos45° + isin45°).
Now, to find the cube roots, we can use De Moivre's Theorem which states that for any complex number in polar form r(cosθ + isinθ), the nth roots are given by the formula:
z^(1/n) = r^(1/n)(cos((θ + 2kπ)/n) + isin((θ + 2kπ)/n))
where k = 0, 1, 2, ..., n-1.
In this case, n = 3 as we are finding cube roots. Therefore, the cube roots are:
1. Cube root 1:
r^(1/3) = 64^(1/3) = 4
θ1 = (45 + 2*0*π)/3 = 15°
Cube root 1 in polar form: 4(cos15° + isin15°)
2. Cube root 2:
r^(1/3) = 64^(1/3) = 4
θ2 = (45 + 2*π)/3 = 105°
Cube root 2 in polar form: 4(cos105° + isin105°)
3. Cube root 3:
r^(1/3) = 64^(1/3) = 4
θ3 = (45 + 2*2*π)/3 = 195°
Cube root 3 in polar form: 4(cos195° + isin195°)
Therefore, the three cube roots of the complex number 64(cos219° + isin219°) in polar form are:
1. 4(cos15° + isin15°)
2. 4(cos105° + isin105°)
3. 4(cos195° + isin195°)
write it like a 12th grader wrote it
First, we converted the complex number 64(cos219° + isin219°) into polar form, which is 64(cos45° + isin45°).
Next, we used De Moivre's Theorem to find the cube roots. The formula for cube roots of a complex number in polar form is z^(1/3) = r^(1/3)(cos((θ + 2kπ)/3) + isin((θ + 2kπ)/3)), where k = 0, 1, 2.
After calculating, we found the cube roots in polar form:
1. 4(cos15° + isin15°)
2. 4(cos105° + isin105°)
3. 4(cos195° + isin195°)
These are the three cube roots of the complex number 64(cos219° + isin219°).