Let \(w = x + iy\), where \(x\) and \(y\) are real numbers. The conjugate \(w^* = x - iy\).

Substitute into the equation:

\((x + iy)^2 + 2i(x - iy) = 1\)

Expand \((x + iy)^2 = x^2 - y^2 + 2ixy\).

Thus, the equation becomes:

\(x^2 - y^2 + 2ixy + 2ix + 2y = 1\)

Equate real and imaginary parts:

Real: \(x^2 - y^2 + 2y = 1\)

Imaginary: \(2xy + 2x = 0\)

From the imaginary part, factor out \(2x\):

\(2x(y + 1) = 0\)

This gives \(x = 0\) or \(y = -1\).

Case 1: \(x = 0\)

Substitute into the real part:

\(-y^2 + 2y = 1\)

\(y^2 - 2y + 1 = 0\)

\((y - 1)^2 = 0\)

\(y = 1\)

Thus, \(w = i\).

Case 2: \(y = -1\)

Substitute into the real part:

\(x^2 - (-1)^2 + 2(-1) = 1\)

\(x^2 - 1 - 2 = 1\)

\(x^2 = 4\)

\(x = \\pm 2\)

Since \(\text{Re} \, w \leq 0\), \(x = -2\).

Thus, \(w = -2 - i\).

The solutions are \(w = -2 - i\) and \(w = i\).