Given that \( z = 1 + i \), let's substitute \( z \) into the original equation and check for consistency:

\[
\frac{a}{1 + (1 + i)} + \frac{b}{1 + (1 + i)^3} = 2i
\]

First, simplify \( 1 + (1 + i) \):
\[
1 + (1 + i) = 2 + i
\]

Next, simplify \( 1 + (1 + i)^3 \):
\[
(1 + i)^3 = (1 + i)(1 + i)(1 + i)
\]
Calculate \( (1 + i)(1 + i) \):
\[
(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i
\]
Then multiply by \( (1 + i) \):
\[
(2i)(1 + i) = 2i + 2i^2 = 2i - 2 = -2 + 2i
\]
Therefore,
\[
1 + (1 + i)^3 = 1 + (-2 + 2i) = -1 + 2i
\]

Now substitute back into the original equation:
\[
\frac{a}{2 + i} + \frac{b}{-1 + 2i} = 2i
\]

To simplify these fractions, multiply the numerator and denominator by the conjugate of the denominator.

For \(\frac{a}{2 + i}\):
\[
\frac{a}{2 + i} \cdot \frac{2 - i}{2 - i} = \frac{a(2 - i)}{(2 + i)(2 - i)} = \frac{a(2 - i)}{4 + 1} = \frac{a(2 - i)}{5} = \frac{2a}{5} - \frac{ai}{5}
\]

For \(\frac{b}{-1 + 2i}\):
\[
\frac{b}{-1 + 2i} \cdot \frac{-1 - 2i}{-1 - 2i} = \frac{b(-1 - 2i)}{(-1 + 2i)(-1 - 2i)} = \frac{b(-1 - 2i)}{1 + 4} = \frac{b(-1 - 2i)}{5} = \frac{-b}{5} - \frac{2bi}{5}
\]

Putting it all together:
\[
\frac{2a}{5} - \frac{ai}{5} + \frac{-b}{5} - \frac{2bi}{5} = 2i
\]

Combine the terms:
\[
\frac{2a - b}{5} + \frac{-a - 2b}{5}i = 2i
\]

For the real and imaginary parts to match, the following system of equations must be true:
\[
\frac{2a - b}{5} = 0 \quad \text{and} \quad \frac{-a - 2b}{5} = 2
\]

Solving the first equation:
\[
2a - b = 0 \implies b = 2a
\]

Solving the second equation:
\[
\frac{-a - 2b}{5} = 2 \implies -a - 2b = 10 \implies -a - 2(2a) = 10 \implies -a - 4a = 10 \implies -5a = 10 \implies a = -2
\]

Substitute \( a = -2 \) into \( b = 2a \):
\[
b = 2(-2) = -4
\]

So the values of \( a \) and \( b \) that satisfy the equation when \( z = 1 + i \) are:
\[
a = -2 \quad \text{and} \quad b = -4
\]

Would you like more details on this solution or have any questions? Here are some further questions you might find interesting:

1. Can you verify the solution for different values of \( a \) and \( b \)?
2. How do the complex conjugates help in simplifying complex fractions?
3. Can we solve the polynomial equation directly for \( z \) without substituting a specific value?
4. What is the geometric interpretation of complex numbers in this context?
5. How would the solution change if \( z \) had a different complex value?

**Tip:** When dealing with complex fractions, multiplying by the conjugate of the denominator is a powerful technique to simplify the expression.

Learn how to solve a complex equation involving z = 1 + i with detailed step-by-step calculations. This problem covers complex numbers, complex fractions, and equations. Suitable for advanced high school students and beyond.

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