Complex Number Identity Proof
Problem Statement
Given \( m \) is a positive integer, show that:
\[ (1 + i)^m + (1 - i)^m = 2^{\frac{m}{2}+1} \cos\left(\frac{m\pi}{4}\right) \]
Proof
Step 1: Express in Polar Form
First, express \( 1 + i \) and \( 1 - i \) in polar form:
\[ 1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) \]
\[ 1 - i = \sqrt{2}\left(\cos\frac{\pi}{4} - i\sin\frac{\pi}{4}\right) \]
Step 2: Apply De Moivre's Theorem
Using De Moivre's Theorem for both expressions:
\[ (1 + i)^m = (\sqrt{2})^m \left(\cos\frac{m\pi}{4} + i\sin\frac{m\pi}{4}\right) \]
\[ (1 - i)^m = (\sqrt{2})^m \left(\cos\frac{m\pi}{4} - i\sin\frac{m\pi}{4}\right) \]
Step 3: Add the Two Expressions
\[ (1 + i)^m + (1 - i)^m = 2^{\frac{m}{2}} \left(2\cos\frac{m\pi}{4}\right) \]
\[ = 2^{\frac{m}{2}+1} \cos\frac{m\pi}{4} \]
Note: The imaginary parts cancel out when adding the two expressions.
Final Result
\[ \boxed{(1 + i)^m + (1 - i)^m = 2^{\frac{m}{2}+1} \cos\left(\frac{m\pi}{4}\right)} \]
Verification for Specific Cases
Case m = 1:
\[ (1+i) + (1-i) = 2 = 2^{\frac{3}{2}}\cos\frac{\pi}{4} = 2\sqrt{2}\cdot\frac{\sqrt{2}}{2} = 2 \]
Case m = 2:
\[ (1+i)^2 + (1-i)^2 = (2i) + (-2i) = 0 = 2^{2}\cos\frac{\pi}{2} = 4\cdot 0 = 0 \]
Case m = 4:
\[ (1+i)^4 + (1-i)^4 = (-4) + (-4) = -8 = 2^{3}\cos\pi = 8\cdot(-1) = -8 \]