Find all the complex solutions of the equation #x^6 + 64 = 0#?
Answer:
#sqrt3 + i, sqrt3-i, -sqrt3 + i, -sqrt3 - i, 2i, -2i#
Answer:
The solutions are
We need
Consequently,
Then,
Then,
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The complex solutions of the equation (x^6 + 64 = 0) are:
[x = \pm \sqrt[6]{-64} = \pm \sqrt[6]{64} \cdot \sqrt[6]{-1} = \pm 2\sqrt[6]{-1} = \pm 2 \left(\cos\left(\frac{\pi}{6} + \frac{2\pi k}{6}\right) + i \sin\left(\frac{\pi}{6} + \frac{2\pi k}{6}\right)\right),]
where (k = 0, 1, 2, 3, 4, 5). Thus, the complex solutions are:
[x = 2\left(\cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right)\right), \quad 2\left(\cos\left(\frac{\pi}{6} + \frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{6} + \frac{\pi}{3}\right)\right), \quad 2\left(\cos\left(\frac{\pi}{6} + \frac{2\pi}{3}\right) + i \sin\left(\frac{\pi}{6} + \frac{2\pi}{3}\right)\right),]
[x = -2\left(\cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right)\right), \quad -2\left(\cos\left(\frac{\pi}{6} + \frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{6} + \frac{\pi}{3}\right)\right), \quad -2\left(\cos\left(\frac{\pi}{6} + \frac{2\pi}{3}\right) + i \sin\left(\frac{\pi}{6} + \frac{2\pi}{3}\right)\right).]
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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