Name: Finding Roots Using De Moivre's Theorem
Uploaded: 2019-07-19
Duration: 9 min 15 s
Description: Let us recall De Moivre's Theorem and use it to find the roots of a given complex equation. Watch this amazing video and learn everything in one go!

Question

If

α, β

are the roots of the equation

x^{2} - 2 x + 4 = 0

, then for any

n \in N

, show that

α^{n} + β^{n} = 2^{n + 1} \cos (\frac{n π}{3})

.

Accepted Answer

Given quadratic equation: $x^{2} - 2 x + 4 = 0$ and to prove: $α^{n} + β^{n} = 2^{n + 1} \cos (\frac{n π}{3})$

$\Rightarrow x = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm 2 i \sqrt{3}}{2} = 1 \pm i \sqrt{3} [∵ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}, when a x^{2} + b x + c = 0]$

Given that, $α, β$ are the roots of the given quadratic equation i.e., $α = 1 + i \sqrt{3} = 2 \{\frac{1}{2} + i \frac{\sqrt{3}}{2}\} = 2 \{\cos \frac{π}{3} + i \sin \frac{π}{3}\}$ and $β = 1 - i \sqrt{3} = 2 \{\frac{1}{2} - i \frac{\sqrt{3}}{2}\} = 2 \{\cos \frac{π}{3} - i \sin \frac{π}{3}\}$

Now, from LHS, we have,

$α^{n} + β^{n} = {\{2 (\cos \frac{π}{3} + i \sin \frac{π}{3})\}}^{n} + {\{2 (\cos \frac{π}{3} - i \sin \frac{π}{3})\}}^{n}$

$= 2^{n} \{\cos \frac{n π}{3} + i \sin \frac{n π}{3} + \cos \frac{n π}{3} - i \sin \frac{n π}{3}\} \{∵ (\cos θ + i \sin θ)^{n} = \cos n θ + i \sin n θ, n \in Z\}$

$= 2^{n} \{2 \cos \frac{n π}{3}\}$

$= 2^{n + 1} \cos \frac{n π}{3}$

$=$ RHS

Hence, proved.

If $α, β$ are the roots of the equation $x^{2} - 2 x + 4 = 0$ , then for any $n \in N$ , show that $α^{n} + β^{n} = 2^{n + 1} \cos (\frac{n π}{3})$ .

Important Questions on De Moivre's Theorem

Learn and Prepare for any exam you want !

If α,β are the roots of the equation x2-2x+4=0, then for any n∈N, show that αn+βn=2n+1cosnπ3.

Important Questions on De Moivre's Theorem

Learn and Prepare for any exam you want !

If $α, β$ are the roots of the equation $x^{2} - 2 x + 4 = 0$ , then for any $n \in N$ , show that $α^{n} + β^{n} = 2^{n + 1} \cos (\frac{n π}{3})$ .