$$\DeclareMathOperator{cosec}{cosec}$$

# Number and Algebra

Syllabus Content

Complex conjugate roots of quadratic and polynomial equations with real coefficients. De Moivre’s theorem and its extension to rational exponents. Powers and roots of complex numbers.

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## Furthermore

Official Guidance, clarification and syllabus links:

Complex roots occur in conjugate pairs.

Includes proof by induction for the case where $$n \in \mathbb{Z^+}$$; awareness that it is true for $$n \in \mathbb{R}$$.

Formula Booklet:
 De Moivre’s theorem $$[r(\cos \theta + i \sin \theta)]^n = r^n (\cos n\theta + i \sin n\theta) = r^n e^{in\theta}=r^n \text{cis} n\theta$$

In the realm of complex numbers, when we deal with quadratic and polynomial equations with real coefficients, an interesting phenomenon occurs: if a complex number is a root, its conjugate is also a root. These pairs are known as complex conjugate roots. Essentially, if $$a + bi$$ is a root, then $$a - bi$$ is also a root, where $$a$$ and $$b$$ are real numbers and $$i$$ is the imaginary unit.

De Moivre's theorem provides a powerful method to compute powers of complex numbers. Stated formally, for any real number $$n$$ and any complex number in polar form $$r(\cos \theta + i \sin \theta)$$, De Moivre's theorem asserts:

$$[r(\cos \theta + i \sin \theta)]^n = r^n (\cos n\theta + i \sin n\theta)$$

This theorem can be extended to rational exponents, allowing us to find both powers and roots of complex numbers with ease. When finding roots, the formula yields multiple distinct values, representing the different roots of the complex number.

Key Formulae:

1. Complex Conjugate: If $$z = a + bi$$ then the conjugate $$\bar{z} = a - bi$$.

2. De Moivre's Theorem:

$$[r(\cos \theta + i \sin \theta)]^n = r^n (\cos n\theta + i \sin n\theta)$$

Finding the Cube Roots of 8:

1. Express the number in its polar form. The number 8 can be expressed as $$8(\cos 0^\circ + i \sin 0^\circ)$$ since it lies on the positive real axis.

2. Apply De Moivre's theorem to find the cube roots. The general formula for the $$n$$-th roots of a complex number $$r(\cos \theta + i \sin \theta)$$ is:

$$\sqrt[n]{r} \left( \cos \left( \frac{\theta + 2k\pi}{n} \right) + i \sin \left( \frac{\theta + 2k\pi}{n} \right) \right)$$

where $$k$$ takes on values from 0 to $$n-1$$.

3. For our case, $$n = 3$$, $$r = 8$$, and $$\theta = 0^\circ$$. Plugging in these values and varying $$k$$ gives:

For $$k = 0$$:

$$\sqrt[3]{8} \left( \cos \left( \frac{0 + 2(0)\pi}{3} \right) + i \sin \left( \frac{0 + 2(0)\pi}{3} \right) \right) = 2(\cos 0^\circ + i \sin 0^\circ)$$

For $$k = 1$$:

$$\sqrt[3]{8} \left( \cos \left( \frac{0 + 2(1)\pi}{3} \right) + i \sin \left( \frac{0 + 2(1)\pi}{3} \right) \right) = 2(\cos 120^\circ + i \sin 120^\circ)$$

For $$k = 2$$:

$$\sqrt[3]{8} \left( \cos \left( \frac{0 + 2(2)\pi}{3} \right) + i \sin \left( \frac{0 + 2(2)\pi}{3} \right) \right) = 2(\cos 240^\circ + i \sin 240^\circ)$$

Thus, the three cube roots of 8 are $$2(\cos 0^\circ + i \sin 0^\circ)$$, $$2(\cos 120^\circ + i \sin 120^\circ)$$, and $$2(\cos 240^\circ + i \sin 240^\circ)$$.

$$= 2, -1 + 1.73205i \text{ and } -1 - 1.73205i$$

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