We are given the equation:

\[
(z+1)(z-1) = i
\]

where \( i \) is the imaginary unit. Let's solve this step by step.

### Step 1: Expand the left-hand side
We start by expanding the product on the left-hand side using the difference of squares formula:

\[
(z+1)(z-1) = z^2 - 1
\]

Thus, the equation becomes:

\[
z^2 - 1 = i
\]

### Step 2: Isolate \( z^2 \)
Now, we isolate \( z^2 \) by adding 1 to both sides:

\[
z^2 = i + 1
\]

### Step 3: Express the right-hand side in polar form
We now express \( i + 1 \) in polar form. The complex number \( 1 + i \) can be written as:

\[
1 + i = \sqrt{2} \text{cis} \left( \frac{\pi}{4} \right)
\]

where \( \text{cis}(\theta) = \cos(\theta) + i\sin(\theta) \), and \( \theta = \frac{\pi}{4} \).

### Step 4: Take the square root of both sides
Now we solve for \( z \) by taking the square root of both sides. Taking the square root of a complex number gives two solutions:

\[
z = \pm \sqrt{\sqrt{2} \text{cis} \left( \frac{\pi}{4} \right)}
\]

Using the property of square roots in polar form:

\[
z = \pm \sqrt[2]{2^{1/2}} \text{cis} \left( \frac{\pi}{8} \right)
\]

This simplifies to:

\[
z = \pm 2^{1/4} \text{cis} \left( \frac{\pi}{8} \right)
\]

Thus, the solutions are:

\[
z = 2^{1/4} \text{cis} \left( \frac{\pi}{8} \right) \quad \text{or} \quad z = -2^{1/4} \text{cis} \left( \frac{\pi}{8} \right)
\]

### Final answer:

The solutions for \( z \) are:

\[
z = 2^{1/4} \text{cis} \left( \frac{\pi}{8} \right) \quad \text{and} \quad z = -2^{1/4} \text{cis} \left( \frac{\pi}{8} \right)
\]

Would you like further details on any part of the process?

Here are some related questions:
1. How do you convert a complex number into polar form?
2. What is the significance of the imaginary unit \(i\)?
3. Can all complex numbers be solved using polar form?
4. What are the applications of solving complex equations like this one?
5. How can we visualize complex solutions geometrically?

**Tip**: When dealing with complex numbers, expressing them in polar form often simplifies operations like multiplication and taking roots.

Learn how to solve the complex equation (z+1)(z-1) = i by expanding the expression and using the polar form of complex numbers. This solution involves key concepts such as the difference of squares, expressing complex numbers in polar form, and solving for square roots of complex numbers. Suitable for students studying complex analysis at the undergraduate level.

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