Complex Number Identity Proof
Problem Statement
If \( n \) is an integer, show that:
\[ (1 + \cosθ + i\sinθ)^n + (1 + \cosθ - i\sinθ)^n = 2^{n+1}\cos^n\left(\frac{θ}{2}\right)\cos\left(\frac{nθ}{2}\right) \]
Proof
Step 1: Rewrite the Expression
Let's rewrite \( 1 + \cosθ \) using a trigonometric identity:
\[ 1 + \cosθ = 2\cos^2\left(\frac{θ}{2}\right) \]
Step 2: Express in Polar Form
Now express the complex numbers in terms of half-angle:
\[ 1 + \cosθ + i\sinθ = 2\cos\left(\frac{θ}{2}\right)\left[\cos\left(\frac{θ}{2}\right) + i\sin\left(\frac{θ}{2}\right)\right] \]
\[ 1 + \cosθ - i\sinθ = 2\cos\left(\frac{θ}{2}\right)\left[\cos\left(\frac{θ}{2}\right) - i\sin\left(\frac{θ}{2}\right)\right] \]
Step 3: Raise to Power n
Using De Moivre's Theorem:
\[ (1 + \cosθ + i\sinθ)^n = 2^n\cos^n\left(\frac{θ}{2}\right)\left[\cos\left(\frac{nθ}{2}\right) + i\sin\left(\frac{nθ}{2}\right)\right] \]
\[ (1 + \cosθ - i\sinθ)^n = 2^n\cos^n\left(\frac{θ}{2}\right)\left[\cos\left(\frac{nθ}{2}\right) - i\sin\left(\frac{nθ}{2}\right)\right] \]
Step 4: Add the Two Expressions
\[ \text{Sum} = 2^n\cos^n\left(\frac{θ}{2}\right)\left[2\cos\left(\frac{nθ}{2}\right)\right] \]
\[ = 2^{n+1}\cos^n\left(\frac{θ}{2}\right)\cos\left(\frac{nθ}{2}\right) \]
Final Result
\[ \boxed{(1 + \cosθ + i\sinθ)^n + (1 + \cosθ - i\sinθ)^n = 2^{n+1}\cos^n\left(\frac{θ}{2}\right)\cos\left(\frac{nθ}{2}\right)} \]
Special Cases Verification
Case n = 1:
\[ \text{LHS} = (1 + \cosθ + i\sinθ) + (1 + \cosθ - i\sinθ) = 2(1 + \cosθ) \]
\[ \text{RHS} = 2^{2}\cos\left(\frac{θ}{2}\right)\cos\left(\frac{θ}{2}\right) = 4\cos^2\left(\frac{θ}{2}\right) = 2(1 + \cosθ) \]
Case θ = 0:
\[ \text{LHS} = (2)^n + (2)^n = 2^{n+1} \]
\[ \text{RHS} = 2^{n+1}\cos^n(0)\cos(0) = 2^{n+1} \]
Case θ = π:
\[ \text{LHS} = (0 + i\sinπ)^n + (0 - i\sinπ)^n = 0 \]
\[ \text{RHS} = 2^{n+1}\cos^n\left(\frac{π}{2}\right)\cos\left(\frac{nπ}{2}\right) = 0 \]