The equation $z^{3}-2z+k=0$, where $k$ is a real constant, has a root $z = 1 + ai$, where $a$ is a positive real constant.

1. State the other root using the conjugate root theorem

Since the polynomial $z^3 - 2z + k = 0$ has purely real coefficients, any non-real roots must occur in complex conjugate pairs.

Given the root $z = 1 + ai$, its complex conjugate is also a root.

Another root is $1 - ai$.

1. Form the quadratic factor of the cubic equation using the two complex conjugate roots.

From part (a), the complex conjugate $1-ai$ is also a root.
The sum of these two roots is $(1+ai) + (1-ai) = 2$.
The product of these two roots is $(1+ai)(1-ai) = 1^2 - (ai)^2 = 1 + a^2$.
The quadratic factor corresponding to these roots is $z^2 - (\text{sum})z + (\text{product}) = z^2 - 2z + (1+a^2)$.

2. Express the cubic equation as a product of the quadratic factor and a linear factor, then expand.

Let the third root be the real number $\alpha$. The cubic polynomial can be factored as:
$z^3 - 2z + k = (z^2 - 2z + 1 + a^2)(z - \alpha)$
Expanding the right hand side gives:
$z^3 - 2z + k = z^3 - \alpha z^2 - 2z^2 + 2\alpha z + (1+a^2)z - \alpha(1+a^2)$
$z^3 - 2z + k = z^3 - (\alpha + 2)z^2 + (2\alpha + 1 + a^2)z - \alpha(1+a^2)$

3. Compare the coefficients of $z^2$ and $z$ to find the third root and the value of $a$.

Comparing the coefficients of $z^2$:
$0 = -(\alpha + 2) \implies \alpha = -2$
Comparing the coefficients of $z$:
$-2 = 2\alpha + 1 + a^2$
Substitute $\alpha = -2$:
$-2 = 2(-2) + 1 + a^2$
$-2 = -3 + a^2 \implies a^2 = 1$
Since $a$ is given as a positive real constant, $a = 1$.

4. Compare the constant terms to find the value of $k$.

Comparing the constant terms:
$k = -\alpha(1+a^2)$
Substitute $\alpha = -2$ and $a = 1$:
$k = -(-2)(1 + 1^2)$
$k = 2(2) = 4$

1. Determine the roots of the equation from the results of earlier parts.

From part (a), the complex roots are $1+ai$ and $1-ai$.
Let the third root be $\gamma$. Since the coefficient of $z^2$ in the cubic equation $z^3-2z+k=0$ is $0$, the sum of the roots is $0$:
$(1+ai) + (1-ai) + \gamma = 0 \implies 2 + \gamma = 0 \implies \gamma = -2$
From part (b), we found $a = 1$.
Thus, the three roots are $1+i$, $1-i$ and $-2$.

2. Find the area of the triangle formed by these roots on the Argand diagram.

On an Argand diagram, the roots correspond to the points $(1,1)$, $(1,-1)$ and $(-2,0)$.
The points $(1,1)$ and $(1,-1)$ lie on a vertical line, forming a base of length:
$\text{Base} = 1 - (-1) = 2$
The perpendicular distance from the third vertex $(-2,0)$ to this vertical line (which is $\text{Re}(z) = 1$) is:
$\text{Height} = 1 - (-2) = 3$
$\text{Area of triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2}(2)(3) = 3$