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:
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For \( k = 0 \): \[ z_0 = e^{i\frac{\pi}{6}} = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + i\frac{1}{2} \]
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For \( k = 1 \): \[ z_1 = e^{i\frac{3\pi}{6}} = e^{i\frac{\pi}{2}} = 0 + i = i \]
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For \( k = 2 \): \[ z_2 = e^{i\frac{5\pi}{6}} = -\cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = -\frac{\sqrt{3}}{2} + i\frac{1}{2} \]
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For \( k = 3 \): \[ z_3 = e^{i\frac{7\pi}{6}} = -\cos\frac{\pi}{6} - i\sin\frac{\pi}{6} = -\frac{\sqrt{3}}{2} - i\frac{1}{2} \]
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For \( k = 4 \): \[ z_4 = e^{i\frac{9\pi}{6}} = e^{i\frac{3\pi}{2}} = 0 - i = -i \]
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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:
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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: \[ (z - \frac{\sqrt{3}}{2} - i\frac{1}{2})(z + \frac{\sqrt{3}}{2} + i\frac{1}{2}) \] \[ = z^2 + (\sqrt{3})(z) + 1 = z^2 - \sqrt{3}z + 1 \]
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For \( z_1 \) and \( z_4 \): \[ (z - i)(z + i) = z^2 + 1 \]
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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:
\[ (z^2 - \sqrt{3}z + 1)(z^2 + 1)(z^2 + \sqrt{3}z + 1) \]