Let $\alpha,\beta$ be the roots of the equation $x^2-\sqrt{6}x+3=0$ such that $\text{Im}(\alpha)>\text{Im}(\beta)$. Let $a,b$ be integers not divisible by 3 and $n$ be a natural number such that $\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^n(a+ib)$, $i=\sqrt{-1}$. Then $n+a+b$ is equal to ______.
Step-by-Step Solution
Key Concept: Roots: $x=\frac{\sqrt{6}\pm i\sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1\pm i)=\sqrt{3}e^{\pm i\pi/4}$. So $\alpha=\sqrt{3}e^{i\pi/4}$, $\beta=\sqrt{3}e^{-i\pi/4}$. Compute $\frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}(\frac{\alpha}{\beta}+1)$ using polar form.
Step 1: Determine the roots $\alpha$ and $\beta$.
The given quadratic equation is $x^2-\sqrt{6}x+3=0$.
Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$$x = \frac{\sqrt{6} \pm \sqrt{(\sqrt{6})^2 - 4(1)(3)}}{2(1)}$$
$$x = \frac{\sqrt{6} \pm \sqrt{6 - 12}}{2}$$
$$x = \frac{\sqrt{6} \pm \sqrt{-6}}{2}$$
$$x = \frac{\sqrt{6} \pm i\sqrt{6}}{2}$$
The roots are $\frac{\sqrt{6}}{2} + i\frac{\sqrt{6}}{2}$ and $\frac{\sqrt{6}}{2} - i\frac{\sqrt{6}}{2}$.
In polar form, $\frac{\sqrt{6}}{2} + i\frac{\sqrt{6}}{2} = \sqrt{\left(\frac{\sqrt{6}}{2}\right)^2 + \left(\frac{\sqrt{6}}{2}\right)^2} \left(\cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right)\right) = \sqrt{\frac{6}{4}+\frac{6}{4}} e^{i\pi/4} = \sqrt{3}e^{i\pi/4}$.
And $\frac{\sqrt{6}}{2} - i\frac{\sqrt{6}}{2} = \sqrt{3}e^{-i\pi/4}$.
Given $\text{Im}(\alpha)>\text{Im}(\beta)$, we have $\alpha = \sqrt{3}e^{i\pi/4}$ and $\beta = \sqrt{3}e^{-i\pi/4}$.
Step 2: Simplify the expression $\frac{\alpha^{99}}{\beta}+\alpha^{98}$.
The expression is $\frac{\alpha^{99}}{\beta}+\alpha^{98}$.
Factor out $\alpha^{98}$:
$$\frac{\alpha^{99}}{\beta}+\alpha^{98} = \alpha^{98}\left(\frac{\alpha}{\beta}+1\right)$$
First, calculate $\frac{\alpha}{\beta}$:
$$\frac{\alpha}{\beta} = \frac{\sqrt{3}e^{i\pi/4}}{\sqrt{3}e^{-i\pi/4}} = e^{i(\pi/4 - (-\pi/4))} = e^{i\pi/2} = i$$
Substitute this into the expression:
$$\alpha^{98}\left(\frac{\alpha}{\beta}+1\right) = \alpha^{98}(i+1)$$
Step 3: Calculate $\alpha^{98}$.
$$\alpha^{98} = (\sqrt{3}e^{i\pi/4})^{98} = (\sqrt{3})^{98} (e^{i\pi/4})^{98} = 3^{98/2} e^{i98\pi/4} = 3^{49} e^{i49\pi/2}$$
Simplify the exponent of $e$:
$$e^{i49\pi/2} = e^{i(24\pi + \pi/2)} = e^{i\pi/2} = i$$
So, $\alpha^{98} = 3^{49}i$.
Step 4: Substitute $\alpha^{98}$ back into the expression.
$$\alpha^{98}(i+1) = (3^{49}i)(1+i)$$
$$= 3^{49}(i+i^2)$$
$$= 3^{49}(i-1)$$
$$= 3^{49}(-1+i)$$
Step 5: Compare with $3^n(a+ib)$ and find $n+a+b$.
We have $3^{49}(-1+i) = 3^n(a+ib)$.
By comparison, $n=49$, $a=-1$, and $b=1$.
The integers $a=-1$ and $b=1$ are not divisible by 3.
The natural number $n=49$ is consistent.
Therefore, $n+a+b = 49 + (-1) + 1 = 49$.
Correct Answer: 49