One of the roots of the equation $z^4 - z^3 - 9z^2 + sz + t = 0$, where $s$ and $t$ are real, is $1 + 2\text{i}$. Find the other roots of the equation and the values of $s$ and $t$. The complex number $w$ is such that $w = a + \text{i}b$, where $a$ and $b$ are positive real numbers. The complex conjugate of $w$ is denoted by $w^*$. Given that $\frac{w^3}{\text{i}w^*}$ is purely imaginary, find $w$ in terms of $a$.

1. State the conjugate root and find the corresponding quadratic factor.

Since the coefficients of the polynomial are real, the complex roots must occur in conjugate pairs. Therefore, another root is $1 - 2\text{i}$.

The quadratic factor corresponding to these conjugate roots is:
$(z - (1 + 2\text{i}))(z - (1 - 2\text{i}))$
$= ((z - 1) - 2\text{i})((z - 1) + 2\text{i})$
$= (z - 1)^2 - (2\text{i})^2$
$= z^2 - 2z + 1 - 4(-1)$
$= z^2 - 2z + 5$

2. Use the quadratic factor to form an identity and find the unknown coefficients.

Let the original polynomial be expressed as the product of the known quadratic factor and an unknown quadratic factor:
$z^4 - z^3 - 9z^2 + sz + t = (z^2 - 2z + 5)(z^2 + az + b)$

Expanding the right side:
$= z^4 + az^3 + bz^2 - 2z^3 - 2az^2 - 2bz + 5z^2 + 5az + 5b$
$= z^4 + (a - 2)z^3 + (b - 2a + 5)z^2 + (5a - 2b)z + 5b$

Comparing the coefficients of $z^3$:
$a - 2 = -1 \implies a = 1$

Comparing the coefficients of $z^2$:
$b - 2a + 5 = -9$
$b - 2(1) + 5 = -9 \implies b + 3 = -9 \implies b = -12$

Thus, the other quadratic factor is $z^2 + z - 12$.

3. Solve for the remaining roots.

To find the remaining roots, set the second quadratic factor to zero:
$z^2 + z - 12 = 0$
$(z + 4)(z - 3) = 0$
$z = -4 \quad \text{or} \quad z = 3$

4. Find the values of $s$ and $t$.

To find the values of $s$ and $t$, compare the remaining coefficients in the identity.

Comparing the coefficients of $z$:
$s = 5a - 2b = 5(1) - 2(-12) = 5 + 24 = 29$

Comparing the constant terms:
$t = 5b = 5(-12) = -60$

1. Expand $w^3$ in terms of $a$ and $b$.

$w = a + \text{i}b$
$w^3 = (a + \text{i}b)^3$
$= a^3 + 3a^2(\text{i}b) + 3a(\text{i}b)^2 + (\text{i}b)^3$
$= a^3 + 3a^2b\text{i} - 3ab^2 - b^3\text{i}$
$= (a^3 - 3ab^2) + \text{i}(3a^2b - b^3)$

2. Simplify the denominator $\text{i}w^*$.

$w^* = a - \text{i}b$
$\text{i}w^* = \text{i}(a - \text{i}b)$
$= a\text{i} - b\text{i}^2 = b + a\text{i}$

3. Express the fraction $\frac{w^3}{\text{i}w^*}$ in Cartesian form by rationalising the denominator.

$\frac{w^3}{\text{i}w^*} = \frac{(a^3 - 3ab^2) + \text{i}(3a^2b - b^3)}{b + a\text{i}}$
Multiply the numerator and denominator by the complex conjugate of the denominator, $b - a\text{i}$:
$= \frac{[(a^3 - 3ab^2) + \text{i}(3a^2b - b^3)](b - a\text{i})}{(b + a\text{i})(b - a\text{i})}$
$= \frac{[(a^3 - 3ab^2)b + (3a^2b - b^3)a] + \text{i}[(3a^2b - b^3)b - (a^3 - 3ab^2)a]}{b^2 + a^2}$
$= \frac{(a^3b - 3ab^3 + 3a^3b - ab^3) + \text{i}(3a^2b^2 - b^4 - a^4 + 3a^2b^2)}{a^2 + b^2}$
$= \frac{4a^3b - 4ab^3}{a^2 + b^2} + \text{i}\frac{6a^2b^2 - a^4 - b^4}{a^2 + b^2}$

4. Use the condition that the complex number is purely imaginary to find $b$ in terms of $a$.

Since the complex number is purely imaginary, its real part must be zero.
$\text{Re}\left(\frac{w^3}{\text{i}w^*}\right) = \frac{4a^3b - 4ab^3}{a^2 + b^2} = 0$
$4ab(a^2 - b^2) = 0$
Since $a$ and $b$ are positive real numbers ($a > 0, b > 0$), $4ab \neq 0$.
$a^2 - b^2 = 0 \implies (a - b)(a + b) = 0$
Since $a + b > 0$, we must have $a - b = 0 \implies b = a$.
Therefore, $w = a + a\text{i}$.