To evaluate

$\lim_{z \to i} \frac{3z^4 - 2z^3 + 8z^2 - 2z + 5}{z - i},$

we notice that directly substituting $z = i$ in the denominator would make it zero, so we need a different approach to resolve this limit.

One effective method is to apply L'Hôpital's Rule, which can be used when a limit is in the form $\frac{0}{0}$. According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately with respect to $z$ until we obtain a determinate form.

Step 1: Differentiate the Numerator and Denominator

1. Numerator: Differentiate $f(z) = 3z^4 - 2z^3 + 8z^2 - 2z + 5$. $f'(z) = 12z^3 - 6z^2 + 16z - 2.$

2. Denominator: Differentiate $g(z) = z - i$. $g'(z) = 1.$

Step 2: Substitute $z = i$

Now we evaluate the limit using these derivatives: $\lim_{z \to i} \frac{3z^4 - 2z^3 + 8z^2 - 2z + 5}{z - i} = \lim_{z \to i} \frac{12z^3 - 6z^2 + 16z - 2}{1}.$

Substituting $z = i$:

1. Since $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$, we can calculate each term:
   • $12z^3 = 12(-i) = -12i$,
   • $-6z^2 = -6(-1) = 6$,
   • $16z = 16i$,
   • $-2$ remains as it is.

Thus, $f'(i) = -12i + 6 + 16i - 2 = (6 - 2) + (-12i + 16i) = 4 + 4i.$

Final Answer

$\lim_{z \to i} \frac{3z^4 - 2z^3 + 8z^2 - 2z + 5}{z - i} = 4 + 4i.$

Would you like additional details, or do you have any questions?

Here are some related questions you might find helpful:

1. How does L'Hôpital's Rule work in cases involving higher derivatives?
2. Could we solve this limit using polynomial division?
3. What are the properties of limits in complex functions?
4. How can we use Taylor expansion to evaluate limits?
5. What are some other methods to evaluate limits with indeterminate forms?

Tip: For limits involving complex functions, converting to polar form or using L'Hôpital's Rule can often simplify the process.