Question

$\left(\frac{1+cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^{8}$

Answer 1

-1

Answer 2

The problem asks us to evaluate the given complex number expression:

$\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^{8}$

Let $z = \cos \frac{\pi}{8} + i \sin \frac{\pi}{8}$. By Euler's formula, $z = e^{i \frac{\pi}{8}}$. Since $z$ is a complex number on the unit circle (i.e., $|z|=1$), its conjugate $\bar{z}$ is equal to $1/z$. So, $\bar{z} = \cos \frac{\pi}{8} - i \sin \frac{\pi}{8} = e^{-i \frac{\pi}{8}}$.

Now, let's rewrite the expression inside the parenthesis using $z$:

$\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}} = \frac{1+\bar{z}}{1+z}$

Since $\bar{z} = 1/z$, we can substitute this into the expression:

$\frac{1+\frac{1}{z}}{1+z} = \frac{\frac{z+1}{z}}{1+z}$

Assuming $1+z \neq 0$ (which is true since $z = e^{i \frac{\pi}{8}} \neq -1$), we can cancel out the $(1+z)$ term:

$\frac{1}{z}$

So, the expression inside the parenthesis simplifies to $\frac{1}{z}$. Substituting back the value of $z$:

$\frac{1}{z} = \frac{1}{\cos \frac{\pi}{8} + i \sin \frac{\pi}{8}} = \cos \frac{\pi}{8} - i \sin \frac{\pi}{8} = e^{-i \frac{\pi}{8}}$

Now, we need to raise this simplified expression to the power of 8:

$\left(e^{-i \frac{\pi}{8}}\right)^{8}$

Using De Moivre's theorem, which states that $(e^{i\theta})^n = e^{in\theta}$:

$e^{-i \frac{\pi}{8} \times 8} = e^{-i\pi}$

Finally, we evaluate $e^{-i\pi}$ using Euler's formula $e^{i\phi} = \cos \phi + i \sin \phi$:

$e^{-i\pi} = \cos(-\pi) + i \sin(-\pi)$

We know that $\cos(-\pi) = \cos(\pi) = -1$ and $\sin(-\pi) = -\sin(\pi) = 0$.

$e^{-i\pi} = -1 + i(0) = -1$

The final answer is -1.

Explanation of the solution:

1. Let $z = \cos \frac{\pi}{8} + i \sin \frac{\pi}{8}$. By Euler's formula, $z = e^{i \frac{\pi}{8}}$.
2. Since $|z|=1$, its conjugate $\bar{z} = \cos \frac{\pi}{8} - i \sin \frac{\pi}{8}$ is equal to $1/z$.
3. Substitute $z$ and $\bar{z}$ into the given expression: $\left(\frac{1+\bar{z}}{1+z}\right)^{8}$.
4. Simplify the term inside the parenthesis: $\frac{1+\bar{z}}{1+z} = \frac{1+1/z}{1+z} = \frac{(z+1)/z}{1+z} = \frac{1}{z}$.
5. So, the expression becomes $\left(\frac{1}{z}\right)^{8}$.
6. Substitute $z = e^{i \frac{\pi}{8}}$ back: $\left(\frac{1}{e^{i \frac{\pi}{8}}}\right)^{8} = \left(e^{-i \frac{\pi}{8}}\right)^{8}$.
7. Apply De Moivre's theorem: $e^{-i \frac{\pi}{8} \times 8} = e^{-i\pi}$.
8. Evaluate $e^{-i\pi} = \cos(-\pi) + i \sin(-\pi) = -1 + 0i = -1$.