Answer :
To prove the statement:
[tex]\[ (\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n + 1} \cos \frac{n \pi}{6} \][/tex]
where [tex]\( n \)[/tex] is a positive integer, let us examine the statement in detail.
Step 1: Express the complex numbers in polar form
First, consider the complex numbers [tex]\(\sqrt{3} + i\)[/tex] and [tex]\(\sqrt{3} - i\)[/tex].
For [tex]\(\sqrt{3} + i\)[/tex]:
- The modulus [tex]\( r \)[/tex] is [tex]\( \sqrt{ (\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \)[/tex].
- The argument [tex]\( \theta \)[/tex] is [tex]\( \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \)[/tex].
Therefore,
[tex]\[ \sqrt{3} + i = 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \][/tex]
Similarly, for [tex]\(\sqrt{3} - i\)[/tex]:
- The modulus is also [tex]\( 2 \)[/tex].
- The argument is [tex]\( -\frac{\pi}{6} \)[/tex].
Thus,
[tex]\[ \sqrt{3} - i = 2 \left( \cos \left( -\frac{\pi}{6} \right) + i \sin \left( -\frac{\pi}{6} \right) \right) \][/tex]
Step 2: Raise these expressions to the [tex]\( n \)[/tex]-th power
Using De Moivre's Theorem which states [tex]\((re^{i \theta})^n = r^n e^{i n\theta}\)[/tex], we have:
[tex]\[ (\sqrt{3} + i)^n = \left[ 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \right]^n = 2^n \left( \cos \frac{n\pi}{6} + i \sin \frac{n\pi}{6} \right) \][/tex]
Similarly,
[tex]\[ (\sqrt{3} - i)^n = \left[ 2 \left( \cos \left(-\frac{\pi}{6}\right) + i \sin \left( -\frac{\pi}{6} \right) \right) \right]^n = 2^n \left( \cos \left( -\frac{n \pi}{6} \right) + i \sin \left( -\frac{n \pi}{6} \right) \right) \][/tex]
Step 3: Add the two expressions
Next, we add [tex]\((\sqrt{3} + i)^n\)[/tex] and [tex]\((\sqrt{3} - i)^n\)[/tex]:
[tex]\[ (\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^n \left( \cos \frac{n \pi}{6} + i \sin \frac{n \pi}{6} \right) + 2^n \left( \cos \left( -\frac{n \pi}{6} \right) + i \sin \left( -\frac{n \pi}{6} \right) \right) \][/tex]
Using the identity that [tex]\(\cos(-x) = \cos(x)\)[/tex] and [tex]\(\sin(-x) = -\sin(x)\)[/tex], we get:
[tex]\[ 2^n \left( \cos \frac{n \pi}{6} + i \sin \frac{n \pi}{6} \right) + 2^n \left( \cos \frac{n \pi}{6} - i \sin \frac{n \pi}{6} \right) \][/tex]
Combining like terms:
[tex]\[ 2^n \left( \cos \frac{n \pi}{6} + \cos \frac{n \pi}{6} \right) = 2 \cdot 2^n \cos \frac{n \pi}{6} \][/tex]
Simplifying:
[tex]\[ 2^{n + 1} \cos \frac{n \pi}{6} \][/tex]
Therefore,
[tex]\[ (\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n + 1} \cos \frac{n \pi}{6} \][/tex]
This completes the proof. The expression holds true for any positive integer [tex]\( n \)[/tex].
[tex]\[ (\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n + 1} \cos \frac{n \pi}{6} \][/tex]
where [tex]\( n \)[/tex] is a positive integer, let us examine the statement in detail.
Step 1: Express the complex numbers in polar form
First, consider the complex numbers [tex]\(\sqrt{3} + i\)[/tex] and [tex]\(\sqrt{3} - i\)[/tex].
For [tex]\(\sqrt{3} + i\)[/tex]:
- The modulus [tex]\( r \)[/tex] is [tex]\( \sqrt{ (\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \)[/tex].
- The argument [tex]\( \theta \)[/tex] is [tex]\( \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \)[/tex].
Therefore,
[tex]\[ \sqrt{3} + i = 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \][/tex]
Similarly, for [tex]\(\sqrt{3} - i\)[/tex]:
- The modulus is also [tex]\( 2 \)[/tex].
- The argument is [tex]\( -\frac{\pi}{6} \)[/tex].
Thus,
[tex]\[ \sqrt{3} - i = 2 \left( \cos \left( -\frac{\pi}{6} \right) + i \sin \left( -\frac{\pi}{6} \right) \right) \][/tex]
Step 2: Raise these expressions to the [tex]\( n \)[/tex]-th power
Using De Moivre's Theorem which states [tex]\((re^{i \theta})^n = r^n e^{i n\theta}\)[/tex], we have:
[tex]\[ (\sqrt{3} + i)^n = \left[ 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \right]^n = 2^n \left( \cos \frac{n\pi}{6} + i \sin \frac{n\pi}{6} \right) \][/tex]
Similarly,
[tex]\[ (\sqrt{3} - i)^n = \left[ 2 \left( \cos \left(-\frac{\pi}{6}\right) + i \sin \left( -\frac{\pi}{6} \right) \right) \right]^n = 2^n \left( \cos \left( -\frac{n \pi}{6} \right) + i \sin \left( -\frac{n \pi}{6} \right) \right) \][/tex]
Step 3: Add the two expressions
Next, we add [tex]\((\sqrt{3} + i)^n\)[/tex] and [tex]\((\sqrt{3} - i)^n\)[/tex]:
[tex]\[ (\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^n \left( \cos \frac{n \pi}{6} + i \sin \frac{n \pi}{6} \right) + 2^n \left( \cos \left( -\frac{n \pi}{6} \right) + i \sin \left( -\frac{n \pi}{6} \right) \right) \][/tex]
Using the identity that [tex]\(\cos(-x) = \cos(x)\)[/tex] and [tex]\(\sin(-x) = -\sin(x)\)[/tex], we get:
[tex]\[ 2^n \left( \cos \frac{n \pi}{6} + i \sin \frac{n \pi}{6} \right) + 2^n \left( \cos \frac{n \pi}{6} - i \sin \frac{n \pi}{6} \right) \][/tex]
Combining like terms:
[tex]\[ 2^n \left( \cos \frac{n \pi}{6} + \cos \frac{n \pi}{6} \right) = 2 \cdot 2^n \cos \frac{n \pi}{6} \][/tex]
Simplifying:
[tex]\[ 2^{n + 1} \cos \frac{n \pi}{6} \][/tex]
Therefore,
[tex]\[ (\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n + 1} \cos \frac{n \pi}{6} \][/tex]
This completes the proof. The expression holds true for any positive integer [tex]\( n \)[/tex].
