(a) Since \(-1 + \sqrt{5}i\) is a root, its complex conjugate \(-1 - \sqrt{5}i\) is also a root because the coefficients of the polynomial are real. Let the third root be \(r\). By Vieta's formulas, the sum of the roots is zero:

\((-1 + \sqrt{5}i) + (-1 - \sqrt{5}i) + r = 0\)

\(-2 + r = 0\)

\(r = 2\)

The polynomial can be expressed as \((z + 1 - \sqrt{5}i)(z + 1 + \sqrt{5}i)(z - 2)\).

Expanding \((z + 1 - \sqrt{5}i)(z + 1 + \sqrt{5}i)\) gives:

\((z + 1)^2 - (\sqrt{5}i)^2 = z^2 + 2z + 1 + 5 = z^2 + 2z + 6\)

Thus, the polynomial is \((z^2 + 2z + 6)(z - 2)\).

Expanding gives:

\(z^3 - 2z^2 + 2z^2 - 4z + 6z - 12 = z^3 + 2z - 12\)

Comparing with \(z^3 + 2z + a = 0\), we find \(a = -12\).

(b) Let \(w = e^{i2\theta}\). Then:

\(\frac{w-1}{w+1} = \frac{e^{i2\theta} - 1}{e^{i2\theta} + 1}\)

Multiply numerator and denominator by \(e^{-i\theta}\):

\(\frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}} = \frac{2i \sin \theta}{2 \cos \theta} = i \tan \theta\)