Exams Joint Entrance Examination Mathematics Complex Numbers Question 69 Question 69 of 455 Medium NUMERICAL
Let α , β \alpha ,\beta α , β be the roots of the equation x 2 − 6 x + 3 = 0 {x}^{2}-\sqrt{6}x+3=0 x 2 − 6 x + 3 = 0 such that I m ( α ) > I m ( β ) Im(\alpha )>Im(\beta ) I m ( α ) > I m ( β ) . Let a , b a,b a , b be integers not divisible by 3 and n n n be a natural number such that α 99 β + α 98 = 3 n ( a + i b ) , i = − 1 \frac{{\alpha }^{99}}{\beta }+{\alpha }^{98}={3}^{n}(a+ib),i=\sqrt{-1} β α 99 + α 98 = 3 n ( a + ib ) , i = − 1 . Then n + a + b n+a+b n + a + b is equal to \_\_\_\_\_\_\_\_\_\_.
Answer: 49
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x = 6 ± i 6 2 = 6 2 ( 1 ± i ) α = 3 ( e i π 4 ) , β = 3 ( e − i π 4 ) ∴ α 99 β + α 98 = α 98 ( α β + 1 ) = α 98 ( α + β ) β = 3 49 ( e i 99 π 4 ) × 2 = 3 49 ( − 1 + i ) = 3 n ( a + i b ) ∴ n = 49 , a = − 1 , b = 1 ∴ n + a + b = 49 − 1 + 1 = 49 \begin{gathered} x=\frac{\sqrt{6}\pm i\sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1\pm i) \\ \alpha =\sqrt{3}({e}^{i\frac{\pi }{4}}),\beta =\sqrt{3}({e}^{-i\frac{\pi }{4}}) \\ \therefore \frac{{\alpha }^{99}}{\beta }+{\alpha }^{98}={\alpha }^{98}(\frac{\alpha }{\beta }+1) \\ =\frac{{\alpha }^{98}(\alpha +\beta )}{\beta }={3}^{49}({e}^{i99\frac{\pi }{4}})\times \sqrt{2} \\ ={3}^{49}(-1+i) \\ ={3}^{n}(a+ib) \\ \therefore n=49,a=-1,b=1 \\ \therefore n+a+b=49-1+1=49 \end{gathered} x = 2 6 ± i 6 = 2 6 ( 1 ± i ) α = 3 ( e i 4 π ) , β = 3 ( e − i 4 π ) ∴ β α 99 + α 98 = α 98 ( β α + 1 ) = β α 98 ( α + β ) = 3 49 ( e i 99 4 π ) × 2 = 3 49 ( − 1 + i ) = 3 n ( a + ib ) ∴ n = 49 , a = − 1 , b = 1 ∴ n + a + b = 49 − 1 + 1 = 49 Answer the question to unlock