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JEE Main 2019 · 9 Jan · Shift I

JEE Main 2019 Mathematics question on Complex Numbers and Quadratic Equations

Let α\alpha and β\beta be two roots of the equation x2+2x+2=0x^2 + 2x + 2 = 0, then α15+β15\alpha^{15} + \beta^{15} is equal to
  1. A.
    256
  2. B.
    512
  3. C.
    -256
  4. D.
    -512
Show answer and solution

Correct option: C

Solution

The roots are α = -1+i and β = -1-i, which in polar form have modulus √2 and argument 3π/4, so α = √2·e^(i·3π/4) and β = √2·e^(-i·3π/4).

Then α15 + β15 = 2·(√2)15·cos(15·3π/4) = 2·2^(15/2)·cos(45π/4).

Since 45π/4 = 11π + π/4, cos(45π/4) = cos(π/4)·cos(11π) = (1/√2)·(-1) = -1/√2.

So α15 + β15 = 2·2^(15/2)·(-1/√2) = -2·2^(15/2)/2^(1/2) = -2·27 = -256.

[JEE Main 2019 · 9 Jan · Shift I]

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