We aim to prove the given equation:

$(1+i)^n + (1-i)^n = 2^{n/2+1} \cos\left(\frac{n\pi}{4}\right).$

Step 1: Represent $1+i$ and $1-i$ in Polar Form

1. The complex number $1+i$ can be written in polar form as: $1+i = \sqrt{2} e^{i\pi/4}.$

2. Similarly, $1-i$ can be written as: $1-i = \sqrt{2} e^{-i\pi/4}.$

Step 2: Raise Both Terms to the Power $n$

Using the polar form and the rule $(re^{i\theta})^n = r^n e^{i n \theta}$, we get:

1. For $(1+i)^n$: $(1+i)^n = (\sqrt{2})^n e^{i n \pi/4} = 2^{n/2} e^{i n \pi/4}.$

2. For $(1-i)^n$: $(1-i)^n = (\sqrt{2})^n e^{-i n \pi/4} = 2^{n/2} e^{-i n \pi/4}.$

Step 3: Add $(1+i)^n$ and $(1-i)^n$

Now, add the two expressions: $(1+i)^n + (1-i)^n = 2^{n/2} e^{i n \pi/4} + 2^{n/2} e^{-i n \pi/4}.$

Factor out $2^{n/2}$: $(1+i)^n + (1-i)^n = 2^{n/2} \left(e^{i n \pi/4} + e^{-i n \pi/4}\right).$

Step 4: Simplify Using Euler's Formula

Using Euler's formula $e^{i\theta} + e^{-i\theta} = 2\cos\theta$, we get: $e^{i n \pi/4} + e^{-i n \pi/4} = 2\cos\left(\frac{n\pi}{4}\right).$

Thus, the sum becomes: $(1+i)^n + (1-i)^n = 2^{n/2} \cdot 2 \cos\left(\frac{n\pi}{4}\right).$

Step 5: Final Expression

Simplify further: $(1+i)^n + (1-i)^n = 2^{n/2+1} \cos\left(\frac{n\pi}{4}\right).$

Conclusion

The equation is proved.

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Do you have any questions or need further clarification? Here are 5 related questions you might find interesting:

1. How can we use Euler's formula to simplify trigonometric expressions involving complex numbers?
2. Why does $(re^{i\theta})^n = r^n e^{i n \theta}$ hold true for complex numbers?
3. What is the geometric significance of $(1+i)^n$ and $(1-i)^n$?
4. How do we derive polar forms of general complex numbers?
5. Can this method be extended to prove other trigonometric identities involving powers of complex numbers?

Tip: Euler's formula is a powerful tool in simplifying expressions involving complex numbers, especially when they include powers and trigonometric terms.