(−4 + 3i) is one root of the equation $az^2 + bz + c = 0$, where $a, b, c \\in \\mathbb{R}$, and $i^2 = -1$.\nWrite the other root.\n\n(b) Use De Moivre’s Theorem to express $(1 + i)^8$ in its simplest form.\n\n(c) $(1 + i)$ is a root of the equation $z^2 + (-2 + i)z + 3 - i = 0$.\nFind its other root in the form $m + ni$, where $m, n \\in \\mathbb{R}$, and $i^2 = -1$. - Leaving Cert Mathematics - Question 1 - 2016

To apply De Moivre’s Theorem, we first express $1 + i$ in polar form. The modulus and argument are calculated as follows:

1. Modulus:
   $r = |1 + i| = \\sqrt{1^2 + 1^2} = \\sqrt{2}$

2. Argument:
   $\theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}$

Thus, we can express $1 + i$ as:

(1+i) = r \\left( \cos \theta + i \sin \theta \right) = \\sqrt{2} \\left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)

Using De Moivre's theorem, we can calculate $(1 + i)^8$ as follows:

$(1 + i)^8 = \left(\sqrt{2}\right)^8 \left(\cos \left(8 \cdot \frac{\pi}{4}\right) + i \sin \left(8 \cdot \frac{\pi}{4}\right)\right)\n= 16 \left(\cos 2\pi + i \sin 2\pi\right)$

Since $\cos 2\pi = 1$ and $\sin 2\pi = 0$, we get:

$(1 + i)^8 = 16 \cdot 1 = 16$

Given the equation $z^2 + (-2 + i)z + (3 - i) = 0$ and one root $z_1 = 1 + i$, we can use Vieta's formulas, which state that the sum of the roots equals the negation of the coefficient of $z$, divided by the coefficient of $z^2$:

Let the other root be $z_2$. Then:

$z_1 + z_2 = -\frac{-2 + i}{1} = 2 - i$

Thus:

$z_2 = (2 - i) - (1 + i)\\ \\ = 2 - i - 1 - i = 1 - 2i$

Therefore, the other root in the form $m + ni$ is:

$z_2 = 1 - 2i$