We are tasked with evaluating the limit:

\[
\lim_{z \to i} \frac{z^2 + 1}{z^6 + 1}
\]

To find the limit, we'll first substitute \( z = i \) into the expression. 

### Step 1: Substituting \( z = i \)

Substitute \( z = i \) in both the numerator and the denominator.

#### Numerator:
\[
z^2 + 1 = i^2 + 1 = (-1) + 1 = 0
\]

#### Denominator:
\[
z^6 + 1 = i^6 + 1
\]

To evaluate \( i^6 \), use the property of powers of \( i \):
\[
i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^5 = i, \quad i^6 = -1
\]

Thus,
\[
i^6 + 1 = (-1) + 1 = 0
\]

### Step 2: Indeterminate form

Both the numerator and denominator equal 0 when \( z = i \), meaning the limit is in the indeterminate form \( \frac{0}{0} \). To resolve this, we'll apply **L'Hopital's Rule**, which is used to evaluate limits of indeterminate forms.

### Step 3: Apply L'Hopital's Rule

L'Hopital's Rule states that for limits of the form \( \frac{0}{0} \), we can differentiate the numerator and the denominator and then take the limit.

First, differentiate the numerator and denominator with respect to \( z \).

- Derivative of the numerator \( z^2 + 1 \) with respect to \( z \):
\[
\frac{d}{dz}(z^2 + 1) = 2z
\]

- Derivative of the denominator \( z^6 + 1 \) with respect to \( z \):
\[
\frac{d}{dz}(z^6 + 1) = 6z^5
\]

Now, the limit becomes:
\[
\lim_{z \to i} \frac{2z}{6z^5}
\]

### Step 4: Simplifying

Simplify the expression:
\[
\lim_{z \to i} \frac{2z}{6z^5} = \lim_{z \to i} \frac{1}{3z^4}
\]

### Step 5: Substitute \( z = i \)

Now, substitute \( z = i \) into the simplified expression:
\[
\frac{1}{3(i)^4} = \frac{1}{3 \cdot 1} = \frac{1}{3}
\]

### Final Answer:
\[
\lim_{z \to i} \frac{z^2 + 1}{z^6 + 1} = \frac{1}{3}
\]

Would you like a more detailed explanation of any step?

Here are some related questions to explore:

1. What is L'Hopital's Rule, and when can it be used?
2. How do the powers of \( i \) cycle, and why is \( i^4 = 1 \)?
3. Can limits be solved without L'Hopital's Rule in indeterminate forms?
4. How do you evaluate complex limits in multivariable contexts?
5. What are other methods to resolve indeterminate forms like \( 0/0 \)?

**Tip:** Always simplify the expression before applying L'Hopital’s Rule to check if a direct substitution can work.

Learn how to evaluate the complex limit lim z→i (z²+1)/(z⁶+1) using L'Hopital's Rule. This step-by-step solution explains the indeterminate form 0/0 and demonstrates differentiation techniques in complex analysis. Suitable for undergraduate students studying calculus and complex numbers.

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