Expand the complex function 1 𝑧 as Taylor series about 𝑧 = 𝑖.

To expand the complex function \( \frac{1}{z} \) as a Taylor series about \( z = i \), we first rewrite the function in a form suitable for series expansion.

### Step 1: Rewrite the function in terms of \( z - i \)

We need to express \( \frac{1}{z} \) in terms of \( z - i \), because we are expanding around \( z = i \).

\[
\frac{1}{z} = \frac{1}{(z - i) + i}
\]

This expression can now be expanded using a geometric series if we can write it in a form like \( \frac{1}{1 - w} \), which has the standard Taylor series expansion:

\[
\frac{1}{1 - w} = 1 + w + w^2 + w^3 + \dots
\]

To achieve this, we factor out \( i \) from the denominator:

\[
\frac{1}{z} = \frac{1}{i \left( 1 + \frac{z - i}{i} \right)} = -\frac{i}{1 + \frac{z - i}{i}}
\]

Now, let \( w = \frac{z - i}{i} \). This gives:

\[
\frac{1}{z} = -i \cdot \frac{1}{1 + w}
\]

### Step 2: Expand using the geometric series

We can now expand \( \frac{1}{1 + w} \) using the geometric series:

\[
\frac{1}{1 + w} = \sum_{n=0}^{\infty} (-w)^n = 1 - w + w^2 - w^3 + \dots
\]

Substituting \( w = \frac{z - i}{i} \), we get:

\[
\frac{1}{z} = -i \left( 1 - \frac{z - i}{i} + \left( \frac{z - i}{i} \right)^2 - \left( \frac{z - i}{i} \right)^3 + \dots \right)
\]

### Step 3: Simplify the series

Now, let's simplify the terms in the series:

\[
\frac{1}{z} = -i \left( 1 - \frac{z - i}{i} + \frac{(z - i)^2}{i^2} - \frac{(z - i)^3}{i^3} + \dots \right)
\]

Since \( i^2 = -1 \), \( i^3 = -i \), and so on, the series simplifies to:

\[
\frac{1}{z} = -i \left( 1 - \frac{z - i}{i} - \frac{(z - i)^2}{1} + \frac{(z - i)^3}{i} + \dots \right)
\]

Finally, multiplying by \( -i \), we get:

\[
\frac{1}{z} = 1 + i(z - i) - (z - i)^2 - i(z - i)^3 + \dots
\]

### Conclusion

The Taylor series expansion of \( \frac{1}{z} \) about \( z = i \) is:

\[
\frac{1}{z} = 1 + i(z - i) - (z - i)^2 - i(z - i)^3 + \dots
\]

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Would you like more details on this process or further clarification?

Here are some related questions:
1. How do you determine the radius of convergence for a Taylor series?
2. Can this series expansion be used to approximate \( \frac{1}{z} \) near \( z = i \)?
3. What is the difference between a Taylor series and a Laurent series?
4. How does the series behave if we expand around other points?
5. How can we use this series to compute values of \( \frac{1}{z} \) near \( z = i \)?

**Tip:** When expanding a function in a Taylor series around a point \( z_0 \), it's useful to express the function in terms of \( z - z_0 \) and use known series expansions, such as the geometric series.

Learn how to expand the complex function 1/z as a Taylor series about z = i. This detailed explanation covers the use of geometric series and the Taylor series expansion formula. Suitable for advanced undergraduate students and beyond.

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