To find the area of the triangle whose vertices are the roots of the equation \( z^3 + iz^2 + 2i = 0 \), we will follow these steps:
### Step 1: Find the roots of the equation
We start by checking if \( z = i \) is a root of the equation:
\[
z^3 + iz^2 + 2i = 0
\]
Substituting \( z = i \):
\[
i^3 + i(i^2) + 2i = -i - 1 + 2i = -1 + i
\]
Since \( -1 + i \neq 0 \), \( z = i \) is not a root. We will use synthetic division to factor the polynomial.
### Step 2: Factor the polynomial
Since \( z = i \) is not a root, we will look for other roots. We can use the Rational Root Theorem or try to find roots by inspection.
Let's factor the polynomial \( z^3 + iz^2 + 2i \) by assuming \( z - i \) is a factor. We perform synthetic division or polynomial long division.
After testing, we find:
\[
z^3 + iz^2 + 2i = (z - i)(z^2 + 2iz - 2)
\]
### Step 3: Solve the quadratic equation
Now we need to solve the quadratic equation \( z^2 + 2iz - 2 = 0 \) using the quadratic formula:
\[
z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1, b = 2i, c = -2 \):
\[
z = \frac{-2i \pm \sqrt{(2i)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}
\]
Calculating the discriminant:
\[
(2i)^2 = -4 \quad \text{and} \quad -4 \cdot 1 \cdot (-2) = 8
\]
So,
\[
z = \frac{-2i \pm \sqrt{-4 + 8}}{2} = \frac{-2i \pm \sqrt{4}}{2} = \frac{-2i \pm 2}{2}
\]
This simplifies to:
\[
z = -i \pm 1
\]
Thus, the roots are:
\[
z_1 = i, \quad z_2 = 1 - i, \quad z_3 = -1 - i
\]
### Step 4: Find the area of the triangle
The vertices of the triangle are \( (0, 1) \), \( (1, -1) \), and \( (-1, -1) \).
Using the formula for the area of a triangle given vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates:
\[
\text{Area} = \frac{1}{2} \left| 0(-1 - (-1)) + 1(-1 - 1) + (-1)(1 - (-1)) \right|
\]
Calculating:
\[
= \frac{1}{2} \left| 0 + 1(-2) + (-1)(2) \right| = \frac{1}{2} \left| -2 - 2 \right| = \frac{1}{2} \left| -4 \right| = \frac{4}{2} = 2
\]
### Final Answer
The area of the triangle is \( 2 \) square units.
