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Let's break down and solve the two problems step by step.

### Problem (8.1)
You are given:
- \( z = \frac{z_1}{z_2} \) where \( z_1 = \tan(\theta) + i \) and \( z_2 = \overline{z_1} \) (the conjugate of \( z_1 \)).
- The task is to find an expression for \( z^n \) with \( n \in \mathbb{N} \).

#### Step 1: Find \( z_2 = \overline{z_1} \)
Since \( z_1 = \tan(\theta) + i \), its conjugate \( z_2 = \overline{z_1} = \tan(\theta) - i \).

#### Step 2: Compute \( z = \frac{z_1}{z_2} \)
Now, we calculate \( z \):
\[
z = \frac{z_1}{z_2} = \frac{\tan(\theta) + i}{\tan(\theta) - i}
\]
To simplify this, multiply both the numerator and denominator by \( \tan(\theta) + i \) (the conjugate of \( z_2 \)):
\[
z = \frac{(\tan(\theta) + i)(\tan(\theta) + i)}{(\tan(\theta) - i)(\tan(\theta) + i)}
\]
The denominator becomes:
\[
(\tan(\theta) - i)(\tan(\theta) + i) = \tan^2(\theta) + 1
\]
The numerator becomes:
\[
(\tan(\theta) + i)^2 = \tan^2(\theta) + 2i\tan(\theta) - 1
\]
Thus, we have:
\[
z = \frac{\tan^2(\theta) + 2i\tan(\theta) - 1}{\tan^2(\theta) + 1}
\]
This expression simplifies to:
\[
z = \frac{\tan^2(\theta) - 1}{\tan^2(\theta) + 1} + i \cdot \frac{2\tan(\theta)}{\tan^2(\theta) + 1}
\]

#### Step 3: General form of \( z^n \)
Let \( z = r e^{i\varphi} \), where \( r \) is the modulus and \( \varphi \) is the argument of \( z \). Then:
\[
z^n = r^n e^{in\

Let z = z1/z2 where z1 = tan(θ) + i and z2 = conjugate(z1). Find an expression for z^n with n ∈ N. Let z = cos(θ) − i(1 + sin(θ)). Determine.

Learn how to express complex numbers in polar form, solve for z^n where z involves tan(θ) and complex conjugates, and understand the relationship between trigonometry and complex numbers using Euler's formula.

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