(a) The quadratic equation is \((2 - i)z^2 + 2z + 2 + i = 0\). Using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2 - i\), \(b = 2\), and \(c = 2 + i\), we find:
\(z = \frac{-2 \pm \sqrt{2^2 - 4(2-i)(2+i)}}{2(2-i)}\)
\(z = \frac{-2 \pm \sqrt{4 - 4((2-i)(2+i))}}{2(2-i)}\)
\(z = \frac{-2 \pm \sqrt{4 - 4(4 + 1)}}{2(2-i)}\)
\(z = \frac{-2 \pm \sqrt{4 - 20}}{2(2-i)}\)
\(z = \frac{-2 \pm \sqrt{-16}}{2(2-i)}\)
\(z = \frac{-2 \pm 4i}{2(2-i)}\)
\(z = \frac{-2 \pm 4i}{4 - 2i}\)
Multiply numerator and denominator by the conjugate of the denominator:
\(z = \frac{(-2 \pm 4i)(2+i)}{(4-2i)(2+i)}\)
\(z = \frac{-4 - 2i \pm 8i + 4i^2}{8 + 2i - 2i - 2i^2}\)
\(z = \frac{-4 - 2i \pm 8i - 4}{8 + 2}\)
\(z = \frac{-8 + 6i}{10}\)
\(z = -\frac{4}{5} + \frac{3}{5}i\)
For the other root:
\(z = \frac{-2 - 4i}{4 - 2i}\)
Multiply numerator and denominator by the conjugate of the denominator:
\(z = \frac{(-2 - 4i)(2+i)}{(4-2i)(2+i)}\)
\(z = \frac{-4 - 8i - 2i - 4i^2}{8 + 2i - 2i - 2i^2}\)
\(z = \frac{-4 - 8i - 2i + 4}{8 + 2}\)
\(z = \frac{-10i}{10}\)
\(z = -i\)
(b) The complex number \(w = 2e^{\frac{1}{4}\pi i}\) has modulus 2 and argument \(\frac{\pi}{4}\). Thus, \(w = 2(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}) = \sqrt{2} + i\sqrt{2}\).
\(w^3 = (2e^{\frac{1}{4}\pi i})^3 = 8e^{\frac{3}{4}\pi i}\) has modulus 8 and argument \(\frac{3\pi}{4}\).
\(w^* = \sqrt{2} - i\sqrt{2}\).
To find the area of triangle \(ABC\), use the formula for the area of a triangle in the complex plane: \(\text{Area} = \frac{1}{2} |\text{Im}(w \cdot \overline{w^3} + w^3 \cdot \overline{w^*} + w^* \cdot \overline{w})|\).
Calculate the area: \(\text{Area} = 10\).
