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Step-by-Step Solution
Step 1: Identify the quadratic equation and its discriminant
The given quadratic equation is
x^2-\sqrt{6}\,x+3=0.
The discriminant is
\Delta = (\sqrt{6})^2 - 4 \cdot 1 \cdot 3 = 6 - 12 = -6.
Since \Delta < 0, the roots are complex and conjugate to each other.
Step 2: Find the roots in standard form
By the quadratic formula, the roots are:
\alpha, \beta = \frac{\sqrt{6} \pm \sqrt{-6}}{2}.
We also know:
1) \alpha + \beta = \sqrt{6} ,
2) \alpha \beta = 3 ,
3) \operatorname{Im}(\alpha) > \operatorname{Im}(\beta).
From these, one can verify that:
\alpha = \frac{\sqrt{6}}{2}(1 + i),
\quad
\beta = \frac{\sqrt{6}}{2}(1 - i),
where \alpha is the one with the positive imaginary part.
Step 3: Express the roots in polar form
First, compute the magnitude of \alpha :
|\alpha|
= \sqrt{\left(\frac{\sqrt{6}}{2}\right)^2 + \left(\frac{\sqrt{6}}{2}\right)^2}
= \sqrt{\frac{6}{4} + \frac{6}{4}}
= \sqrt{\frac{12}{4}}
= \sqrt{3}.
Since the real and imaginary parts of \alpha are equal and positive, its argument is \frac{\pi}{4}. Hence,
\alpha = \sqrt{3}\,e^{i\,\frac{\pi}{4}},
\quad
\beta = \sqrt{3}\,e^{-\,i\,\frac{\pi}{4}}.
Step 4: Compute \alpha^{98}
Raise \alpha to the 98th power:
\alpha^{98}
= \left(\sqrt{3}\,e^{i\,\frac{\pi}{4}}\right)^{98}
= (\sqrt{3})^{98} \, e^{i\,\frac{98\pi}{4}}
= 3^{49} \, e^{i\,\frac{98\pi}{4}}.
Notice that
\frac{98\pi}{4} = 24.5\,\pi = 24\pi + \frac{\pi}{2}.
Since e^{i\,24\pi} = 1, we get
\alpha^{98} = 3^{49}\,e^{i\,\frac{\pi}{2}} = 3^{49}\,i.
Step 5: Compute \frac{\alpha^{99}}{\beta}
First, find \alpha^{99} :
\alpha^{99} = \alpha^{98} \cdot \alpha = 3^{49}\,i \times \sqrt{3}\,e^{i\,\frac{\pi}{4}}
= 3^{49}\,\sqrt{3}\,i \, e^{i\,\frac{\pi}{4}}.
Recall that i = e^{i\,\frac{\pi}{2}}. Therefore,
\alpha^{99}
= 3^{49}\,\sqrt{3}\, e^{i\,\left(\frac{\pi}{2} + \frac{\pi}{4}\right)}
= 3^{49}\,\sqrt{3}\, e^{i\,\frac{3\pi}{4}}.
Now divide by \beta :
\frac{\alpha^{99}}{\beta}
= \frac{3^{49}\,\sqrt{3}\, e^{i\,\frac{3\pi}{4}}}{\sqrt{3}\, e^{-\,i\,\frac{\pi}{4}}}
= 3^{49}\,e^{i\,\left(\frac{3\pi}{4} + \frac{\pi}{4}\right)}
= 3^{49}\,e^{i\,\pi}
= 3^{49}\,(-1)
= -\,3^{49}.
Step 6: Form the given expression
We need
\frac{\alpha^{99}}{\beta} + \alpha^{98}.
Substituting the values found:
\frac{\alpha^{99}}{\beta} + \alpha^{98}
= (-\,3^{49}) + (3^{49}\,i)
= 3^{49}\,(i - 1).
In the form 3^n(a + i\,b), this becomes
3^{49}\Bigl((-1) + i \cdot 1\Bigr),
so n = 49 , a = -1 , and b = 1.
Step 7: Evaluate n + a + b
Finally,
n + a + b = 49 + (-1) + 1 = 49.
Therefore, the required value is \boxed{49}.
