To solve the problem of finding [tex]\( z^n + \frac{1}{z^n} \)[/tex] where [tex]\( z = \cos \theta + i \sin \theta \)[/tex], let's proceed step by step:
### Step 1: Write down [tex]\( z \)[/tex] in exponential form
We know that the expression [tex]\( \cos \theta + i \sin \theta \)[/tex] can be written using Euler's formula as:
[tex]\[ z = e^{i\theta} \][/tex]
This means:
[tex]\[ z^n = (e^{i\theta})^n = e^{in\theta} \][/tex]
### Step 2: Calculate [tex]\( \frac{1}{z^n} \)[/tex]
To find [tex]\( \frac{1}{z^n} \)[/tex], use the property of exponents:
[tex]\[ \frac{1}{z^n} = \frac{1}{e^{in\theta}} = e^{-in\theta} \][/tex]
### Step 3: Add [tex]\( z^n \)[/tex] and [tex]\( \frac{1}{z^n} \)[/tex]
Now we combine these two results:
[tex]\[ z^n + \frac{1}{z^n} = e^{in\theta} + e^{-in\theta} \][/tex]
### Step 4: Use the cosine addition formula
Recall the formula for the sum of exponentials, which relates to the cosine function:
[tex]\[ e^{ix} + e^{-ix} = 2\cos(x) \][/tex]
Applying this to our expression:
[tex]\[ z^n + \frac{1}{z^n} = e^{in\theta} + e^{-in\theta} = 2\cos(n\theta) \][/tex]
### Conclusion
Thus, the value of [tex]\( z^n + \frac{1}{z^n} \)[/tex] is given by:
[tex]\[ z^n + \frac{1}{z^n} = 2 \cos (n \theta) \][/tex]
By following these steps carefully, you can solve the problem and find the required expression.
