Explanation
\begin{array}{l}
\alpha \text{ \& } \beta \text{ are roots of eq}^n \\
\quad x^2 - \sqrt{6}x + 3 = 0 \\
\begin{aligned}
x &= \frac{\sqrt{6} \pm \sqrt{6-12}}{2} \\
&= \frac{\sqrt{6} \pm \sqrt{6}i}{2} \\
&= \frac{\sqrt{6}}{2}(1 \pm i)
\end{aligned} \\
\text{Im}(\alpha) > \text{Im}(\beta) \\
\alpha = \frac{\sqrt{6}}{2}(1+i) \\
\beta = \frac{\sqrt{6}}{2}(1-i) \\
\begin{aligned}
\alpha &= re^{i\theta} \\
&= \sqrt{3}e^{i \frac{\pi}{4}}
\end{aligned} \\
\alpha^{98} = (\sqrt{3})^{98} \left( e^{i 98 \times \frac{\pi}{4}} \right)
\end{array}
\begin{aligned}
&= 3^{49} \left[ e^{i 49 \frac{\pi}{2}} \right] \\
&= 3^{49} \left( 0 + i \sin 49 \frac{\pi}{2} \right) \\
&= i(3^{49}) \\
\therefore \quad \frac{\alpha^{99}}{\beta} + \alpha^{98} &= \alpha^{98} \left( \frac{\alpha}{\beta} + 1 \right) \\
\Rightarrow i \left( 3^{49} \left( \frac{1 + i}{1 - i} + 1 \right) \right) \\
\Rightarrow 3^{49} \left( \frac{i - 1}{1 - i} + i \right) \\
&= 3^{49} (-1 + i) \\
\Rightarrow \quad a &= -1, \\
b &= 1 \\
n &= 49 \\
n + a + b &= 49
\end{aligned}
