Question

Find the roots of z^6 + 1 and hence resolve z^6 + 1 into real quadratic factors

Answer 1

To find the roots of \( z^6 + 1 = 0 \), we can first rewrite the equation as:

\[
z^6 = -1
\]

Now, we express \(-1\) in polar form. In the complex plane, \(-1\) can be represented as:

\[
-1 = e^{i\pi}
\]

However, it can also be expressed with a periodic nature due to the properties of complex exponents. Therefore, we can express \(-1\) as:

\[
-1 = e^{i(\pi + 2k\pi)} \quad \text{for any integer } k
\]

To find the sixth roots of \(-1\), we set:

\[
z = e^{i(\frac{\pi + 2k\pi}{6})}
\]

for \( k = 0, 1, 2, 3, 4, 5 \). This gives us the six roots:

1. For \( k = 0 \):
\[
z_0 = e^{i\frac{\pi}{6}} = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + i\frac{1}{2}
\]

2. For \( k = 1 \):
\[
z_1 = e^{i\frac{3\pi}{6}} = e^{i\frac{\pi}{2}} = 0 + i = i
\]

3. For \( k = 2 \):
\[
z_2 = e^{i\frac{5\pi}{6}} = -\cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = -\frac{\sqrt{3}}{2} + i\frac{1}{2}
\]

4. For \( k = 3 \):
\[
z_3 = e^{i\frac{7\pi}{6}} = -\cos\frac{\pi}{6} - i\sin\frac{\pi}{6} = -\frac{\sqrt{3}}{2} - i\frac{1}{2}
\]

5. For \( k = 4 \):
\[
z_4 = e^{i\frac{9\pi}{6}} = e^{i\frac{3\pi}{2}} = 0 - i = -i
\]

6. For \( k = 5 \):
\[
z_5 = e^{i\frac{11\pi}{6}} = \cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6} = \frac{\sqrt{3}}{2} - i\frac{1}{2}
\]

Thus, the six roots of \( z^6 + 1 = 0 \) are:

\[
z_0 = \frac{\sqrt{3}}{2} + i\frac{1}{2}, \quad z_1 = i, \quad z_2 = -\frac{\sqrt{3}}{2} + i\frac{1}{2}, \quad z_3 = -\frac{\sqrt{3}}{2} - i\frac{1}{2}, \quad z_4 = -i, \quad z_5 = \frac{\sqrt{3}}{2} - i\frac{1}{2}
\]

Next, we can resolve \( z^6 + 1 \) into real quadratic factors. The roots can be grouped as follows:

- \( z_0 \) and \( z_3 \);
- \( z_1 \) and \( z_4 \);
- \( z_2 \) and \( z_5 \).

We can form quadratic factors from those pairs:

1. **For \( z_0 \) and \( z_3 \):**
\[
(z - z_0)(z - z_3) = (z - (\frac{\sqrt{3}}{2} + i\frac{1}{2}))(z - (-\frac{\sqrt{3}}{2} - i\frac{1}{2}))
\]

This simplifies to:
\[
(z - \frac{\sqrt{3}}{2} - i\frac{1}{2})(z + \frac{\sqrt{3}}{2} + i\frac{1}{2})
\]
\[
= z^2 + (\sqrt{3})(z) + 1 = z^2 - \sqrt{3}z + 1
\]

2. **For \( z_1 \) and \( z_4 \):**
\[
(z - i)(z + i) = z^2 + 1
\]

3. **For \( z_2 \) and \( z_5 \):**
\[
(z - (-\frac{\sqrt{3}}{2} + i\frac{1}{2}))(z - (\frac{\sqrt{3}}{2} - i\frac{1}{2}))
\]
This gives:
\[
= z^2 + (\sqrt{3})(z) + 1 = z^2 + \sqrt{3}z + 1
\]

Now we can write \( z^6 + 1 \) in its factored form:

\[
z^6 + 1 = (z^2 - \sqrt{3}z + 1)(z^2 + 1)(z^2 + \sqrt{3}z + 1)
\]

Thus, the factorization of \( z^6 + 1 \) into real quadratic factors is:

\[
(z^2 - \sqrt{3}z + 1)(z^2 + 1)(z^2 + \sqrt{3}z + 1)
\]