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Question: Let, alpha and beta be two roots of the equation x^2 +2x+2=0,then (alpha)^15 +(beta)^15 is equal to?...

Let, alpha and beta be two roots of the equation x^2 +2x+2=0,then (alpha)^15 +(beta)^15 is equal to?

A

-256

B

256

C

0

D

128

Answer

-256

Explanation

Solution

The given quadratic equation is x2+2x+2=0x^2 + 2x + 2 = 0. The roots are found using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. For this equation, a=1a=1, b=2b=2, c=2c=2. x=−2±22−4(1)(2)2(1)=−2±4−82=−2±−42=−2±2i2=−1±ix = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = \frac{-2 \pm 2i}{2} = -1 \pm i. Let the roots be α=−1+i\alpha = -1 + i and β=−1−i\beta = -1 - i.

We need to find α15+β15\alpha^{15} + \beta^{15}.

Method 1: Using Polar Form and De Moivre's Theorem

Convert the roots to polar form z=r(cos⁡θ+isin⁡θ)=reiθz = r(\cos \theta + i \sin \theta) = re^{i\theta}.

For α=−1+i\alpha = -1 + i: Magnitude r=∣α∣=(−1)2+12=2r = |\alpha| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}. The argument θ\theta is in the second quadrant. tan⁡θ=1−1=−1\tan \theta = \frac{1}{-1} = -1. The principal argument is θ=3π4\theta = \frac{3\pi}{4}. So, α=2(cos⁡(3π4)+isin⁡(3π4))=2ei3π4\alpha = \sqrt{2} \left(\cos\left(\frac{3\pi}{4}\right) + i \sin\left(\frac{3\pi}{4}\right)\right) = \sqrt{2} e^{i \frac{3\pi}{4}}.

For β=−1−i\beta = -1 - i: Magnitude r=∣β∣=(−1)2+(−1)2=2r = |\beta| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}. The argument θ\theta is in the third quadrant. tan⁡θ=−1−1=1\tan \theta = \frac{-1}{-1} = 1. The principal argument is θ=−3π4\theta = -\frac{3\pi}{4}. So, β=2(cos⁡(−3π4)+isin⁡(−3π4))=2e−i3π4\beta = \sqrt{2} \left(\cos\left(-\frac{3\pi}{4}\right) + i \sin\left(-\frac{3\pi}{4}\right)\right) = \sqrt{2} e^{-i \frac{3\pi}{4}}.

Applying De Moivre's theorem: α15=(2)15(cos⁡(15×3π4)+isin⁡(15×3π4))=215/2(cos⁡(45π4)+isin⁡(45π4))\alpha^{15} = (\sqrt{2})^{15} \left(\cos\left(15 \times \frac{3\pi}{4}\right) + i \sin\left(15 \times \frac{3\pi}{4}\right)\right) = 2^{15/2} \left(\cos\left(\frac{45\pi}{4}\right) + i \sin\left(\frac{45\pi}{4}\right)\right). Simplify the angle: 45π4=11π+π4\frac{45\pi}{4} = 11\pi + \frac{\pi}{4}. cos⁡(45π4)=cos⁡(11π+π4)=−cos⁡(π4)=−12\cos\left(\frac{45\pi}{4}\right) = \cos\left(11\pi + \frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}. sin⁡(45π4)=sin⁡(11π+π4)=−sin⁡(π4)=−12\sin\left(\frac{45\pi}{4}\right) = \sin\left(11\pi + \frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}. Also, 215/2=27⋅21/2=12822^{15/2} = 2^7 \cdot 2^{1/2} = 128\sqrt{2}. So, α15=1282(−12−i12)=−128(1+i)\alpha^{15} = 128\sqrt{2} \left(-\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right) = -128(1+i).

For β15\beta^{15}: β15=(2)15(cos⁡(15×−3π4)+isin⁡(15×−3π4))=215/2(cos⁡(−45π4)+isin⁡(−45π4))\beta^{15} = (\sqrt{2})^{15} \left(\cos\left(15 \times -\frac{3\pi}{4}\right) + i \sin\left(15 \times -\frac{3\pi}{4}\right)\right) = 2^{15/2} \left(\cos\left(-\frac{45\pi}{4}\right) + i \sin\left(-\frac{45\pi}{4}\right)\right). cos⁡(−45π4)=cos⁡(45π4)=−12\cos\left(-\frac{45\pi}{4}\right) = \cos\left(\frac{45\pi}{4}\right) = -\frac{1}{\sqrt{2}}. sin⁡(−45π4)=−sin⁡(45π4)=12\sin\left(-\frac{45\pi}{4}\right) = -\sin\left(\frac{45\pi}{4}\right) = \frac{1}{\sqrt{2}}. So, β15=1282(−12+i12)=128(−1+i)\beta^{15} = 128\sqrt{2} \left(-\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right) = 128(-1+i).

Summing them: α15+β15=−128(1+i)+128(−1+i)=(−128−128i)+(−128+128i)=−256\alpha^{15} + \beta^{15} = -128(1+i) + 128(-1+i) = (-128 - 128i) + (-128 + 128i) = -256.

Method 3: Using Conjugate Property Since β=−1−i\beta = -1 - i is the complex conjugate of α=−1+i\alpha = -1 + i, we have β=αˉ\beta = \bar{\alpha}. Therefore, β15=(αˉ)15=α15‾\beta^{15} = (\bar{\alpha})^{15} = \overline{\alpha^{15}}. So, α15+β15=α15+α15‾=2Re(α15)\alpha^{15} + \beta^{15} = \alpha^{15} + \overline{\alpha^{15}} = 2 \text{Re}(\alpha^{15}). From Method 1, α15=−128−128i\alpha^{15} = -128 - 128i. The real part is Re(α15)=−128\text{Re}(\alpha^{15}) = -128. Thus, α15+β15=2×(−128)=−256\alpha^{15} + \beta^{15} = 2 \times (-128) = -256.