Let \alpha, \beta be the roots of the equation x^2-\sqrt6 x+3=0 such that \operatornameIm(\alpha)\gt \operatornameIm(\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+i b), i=\sqrt-1 . Then n+a+b is equal to \underline\hspace2cm .

Complex Numbers PYQ — JEE Main 2024 · 29 Jan · Shift 2 · Q55 with Solution

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Correct answer: 49

Find the roots $\alpha,\beta$

Given $x^2-\sqrt{6}x+3=0.$

Discriminant: $D=(\sqrt{6})^2-4\cdot 3=6-12=-6.$ So the roots are $x=\frac{\sqrt{6}\pm i\sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1\pm i).$

Since $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$ , $\alpha=\frac{\sqrt{6}}{2}(1+i), \qquad \beta=\frac{\sqrt{6}}{2}(1-i).$

Now write them in polar form. Since $1+i=\sqrt{2}\,\operatorname{cis}\left(\frac\pi4\right), \qquad 1-i=\sqrt{2}\,\operatorname{cis}\left(-\frac\pi4\right),$ we get $\alpha=\frac{\sqrt{6}}{2}\cdot \sqrt{2}\,\operatorname{cis}\left(\frac\pi4\right)=\sqrt{3}\,\operatorname{cis}\left(\frac\pi4\right),$ $\beta=\sqrt{3}\,\operatorname{cis}\left(-\frac\pi4\right).$

Also note that $\beta=\overline{\alpha}$ and $\alpha\beta=3.$ Hence $\frac1\beta=\frac{\alpha}{3}.$

Simplify the given expression

We need to evaluate $\frac{\alpha^{99}}{\beta}+\alpha^{98}.$

Using $\dfrac1\beta=\dfrac\alpha3$ , $\frac{\alpha^{99}}{\beta}=\alpha^{99}\cdot \frac\alpha3=\frac{\alpha^{100}}{3}.$ So

=\alpha^{98}\left(\frac{\alpha^2}{3}+1\right).$$ Now compute $\alpha^2$: $$\alpha=\sqrt{3}\,\operatorname{cis}\left(\frac\pi4\right) \implies \alpha^2=3\,\operatorname{cis}\left(\frac\pi2\right)=3i.$$ Therefore $$\frac{\alpha^2}{3}=i,$$ and hence $$\frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}(1+i).$$ --- 3. **Compute $\alpha^{98}$** Since $$\alpha=\sqrt{3}\,\operatorname{cis}\left(\frac\pi4\right),$$ we have $$\alpha^{98}=(\sqrt{3})^{98}\operatorname{cis}\left(\frac{98\pi}{4}\right) =3^{49}\operatorname{cis}\left(\frac{49\pi}{2}\right).$$ Now $$\frac{49\pi}{2}=24\pi+\frac\pi2,$$ so $$\operatorname{cis}\left(\frac{49\pi}{2}\right)=\operatorname{cis}\left(\frac\pi2\right)=i.$$ Thus $$\alpha^{98}=3^{49}i.$$ Therefore, $$\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^{49}i(1+i)=3^{49}(i+i^2)=3^{49}(-1+i).$$ So comparing with $$3^n(a+ib),$$ we get $$n=49, \quad a=-1, \quad b=1.$$ These are integers and neither is divisible by $3$, as required. --- 4. **Find $n+a+b$** $$n+a+b=49+(-1)+1=49.$$ --- 5. **Comparison with stored answer** Derived answer: $49$. Stored correct answer: $49$. They match.

More from Complex Numbers

View written solutionFree

Correct answer: 49

Find the roots $\alpha,\beta$

Given $x^2-\sqrt{6}x+3=0.$

Discriminant: $D=(\sqrt{6})^2-4\cdot 3=6-12=-6.$ So the roots are $x=\frac{\sqrt{6}\pm i\sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1\pm i).$

Since $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$ , $\alpha=\frac{\sqrt{6}}{2}(1+i), \qquad \beta=\frac{\sqrt{6}}{2}(1-i).$

Also note that $\beta=\overline{\alpha}$ and $\alpha\beta=3.$ Hence $\frac1\beta=\frac{\alpha}{3}.$

Simplify the given expression

We need to evaluate $\frac{\alpha^{99}}{\beta}+\alpha^{98}.$

Using $\dfrac1\beta=\dfrac\alpha3$ , $\frac{\alpha^{99}}{\beta}=\alpha^{99}\cdot \frac\alpha3=\frac{\alpha^{100}}{3}.$ So

=\alpha^{98}\left(\frac{\alpha^2}{3}+1\right).$$ Now compute $\alpha^2$: $$\alpha=\sqrt{3}\,\operatorname{cis}\left(\frac\pi4\right) \implies \alpha^2=3\,\operatorname{cis}\left(\frac\pi2\right)=3i.$$ Therefore $$\frac{\alpha^2}{3}=i,$$ and hence $$\frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}(1+i).$$ --- 3. **Compute $\alpha^{98}$** Since $$\alpha=\sqrt{3}\,\operatorname{cis}\left(\frac\pi4\right),$$ we have $$\alpha^{98}=(\sqrt{3})^{98}\operatorname{cis}\left(\frac{98\pi}{4}\right) =3^{49}\operatorname{cis}\left(\frac{49\pi}{2}\right).$$ Now $$\frac{49\pi}{2}=24\pi+\frac\pi2,$$ so $$\operatorname{cis}\left(\frac{49\pi}{2}\right)=\operatorname{cis}\left(\frac\pi2\right)=i.$$ Thus $$\alpha^{98}=3^{49}i.$$ Therefore, $$\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^{49}i(1+i)=3^{49}(i+i^2)=3^{49}(-1+i).$$ So comparing with $$3^n(a+ib),$$ we get $$n=49, \quad a=-1, \quad b=1.$$ These are integers and neither is divisible by $3$, as required. --- 4. **Find $n+a+b$** $$n+a+b=49+(-1)+1=49.$$ --- 5. **Comparison with stored answer** Derived answer: $49$. Stored correct answer: $49$. They match.

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