Question

The roots of ${x^2} + 2x + 2$ are $\alpha $ and $\beta $ . What is the value of ${\alpha ^{15}} + {\beta ^{15}}$ ?

Answer 1

Hint: Any complex number, in its exponential form, can be represented as \[z = r\left( {\cos \theta + i\sin \theta } \right) = r{e^{i\theta }}\] , where $r$ is its magnitude and $\theta $ is its argument. This is said to be the Euler’s form of the complex number, expressed in the terms of its polar coordinates. Using the exponential form, any power of a complex number can be calculated as ${z^n} = r\left( {\cos n\theta + i\sin n\theta } \right) = r{e^{in\theta }}$ .

Formula Used: Using the exponential form, any power of a complex number can be calculated as ${z^n} = r\left( {\cos n\theta + i\sin n\theta } \right) = r{e^{in\theta }}$ .

Complete step by step Solution:
Given quadratic equation: ${x^2} + 2x + 2$
Its roots are given by:
$x = \dfrac{{ - 2 \pm \sqrt {4 - 4(2)} }}{2}$
Simplifying further,
$x = - 1 \pm i$
Let $\alpha = - 1 + i$ and $\beta = - 1 - i$ .
Writing $\alpha $ in its polar form, we have
$\alpha = \sqrt 2 \left( {\cos \dfrac{{3\pi }}{4} + i\sin \dfrac{{3\pi }}{4}} \right)$ … (1)
Similarly, writing $\beta $ in its polar form, we have
$\beta = \sqrt 2 \left( {\cos \dfrac{{5\pi }}{4} + i\sin \dfrac{{5\pi }}{4}} \right)$ … (2)
Now, Euler’s form of a complex number is given as:
$z = r(\cos \theta + i\sin \theta ) = r{e^{i\theta }}$
Hence, on writing complex numbers $\alpha $ and $\beta $ in their Euler’s forms, we get:
$\alpha = \sqrt 2 {e^{i\dfrac{{3\pi }}{4}}}$ … (3)
And
$\beta = \sqrt 2 {e^{i\dfrac{{5\pi }}{4}}}$ … (4)
Taking the power of 15 of equations (3) and (4) and adding them together,
${\alpha ^{15}} + {\beta ^{15}} = {\left( {\sqrt 2 } \right)^{15}}\left( {{e^{i\dfrac{{45\pi }}{4}}} + {e^{i\dfrac{{75\pi }}{4}}}} \right)$
As $\dfrac{{45\pi }}{4} = \dfrac{{3\pi }}{4}$ and $\dfrac{{75\pi }}{4} = \dfrac{{5\pi }}{4}$ , therefore simplifying the above equation, we get:
${\alpha ^{15}} + {\beta ^{15}} = {\left( {\sqrt 2 } \right)^{15}}\left( {{e^{i\dfrac{{3\pi }}{4}}} + {e^{i\dfrac{{5\pi }}{4}}}} \right)$
Using equations (3) and (4), we get:
\[{\alpha ^{15}} + {\beta ^{15}} = {\left( {\sqrt 2 } \right)^{15}}\dfrac{{\left( {\alpha + \beta } \right)}}{{\sqrt 2 }}\]
We also know that $\alpha = - 1 + i$ and $\beta = - 1 - i$ , hence substituting these values in the above equation, we get:
\[{\alpha ^{15}} + {\beta ^{15}} = {\left( {\sqrt 2 } \right)^{15}}\dfrac{{\left( { - 1 + i - 1 - i} \right)}}{{\sqrt 2 }}\]
Simplifying further,
\[{\alpha ^{15}} + {\beta ^{15}} = {\left( {\sqrt 2 } \right)^{14}}\left( { - 2} \right) = - {(2)^8}\]
This gives us:
\[{\alpha ^{15}} + {\beta ^{15}} = - 256\]

Note: Euler’s formula is one of the most important properties of complex numbers as it relates the trigonometric functions along with the exponential function. The above can question can be solved with normal calculations, that is, without using Euler’s form as well.