If the complex number `omega=x+iy(AA x, y in R and i^(2)=-1)` satisfy the equation `omega^(3)=8i`, then the maximum vlaue of y is

Options

A. 1
B. `(sqrt3)/(2)`
C. `(1)/(2)`
D. 2

Answer

To find the maximum value of \( y \) for the complex number \( \omega = x + iy \) satisfying the equation \( \omega^3 = 8i \), we will follow these steps:

### Step 1: Express \( \omega^3 \)
We start with the expression for \( \omega \):
\[ \omega = x + iy \]
Now, we need to compute \( \omega^3 \):
\[ \omega^3 = (x + iy)^3 \]
Using the binomial expansion, we have:
\[ \omega^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 \]
Calculating each term:
- \( (iy)^2 = -y^2 \)
- \( (iy)^3 = -iy^3 \)

Thus, we can rewrite \( \omega^3 \):
\[ \omega^3 = x^3 + 3x^2(iy) - 3xy^2 - iy^3 \]
Combining the real and imaginary parts, we get:
\[ \omega^3 = (x^3 - 3xy^2) + i(3x^2y - y^3) \]

### Step 2: Set \( \omega^3 \) equal to \( 8i \)
Given that \( \omega^3 = 8i \), we can equate the real and imaginary parts:
\[ x^3 - 3xy^2 = 0 \quad \text{(Real part)} \]
\[ 3x^2y - y^3 = 8 \quad \text{(Imaginary part)} \]

### Step 3: Solve the real part equation
From the real part equation:
\[ x^3 - 3xy^2 = 0 \]
We can factor this:
\[ x(x^2 - 3y^2) = 0 \]
This gives us two cases:
1. \( x = 0 \)
2. \( x^2 = 3y^2 \)

We will consider the second case, \( x^2 = 3y^2 \).

### Step 4: Substitute \( x \) in the imaginary part equation
From \( x^2 = 3y^2 \), we find \( x = \sqrt{3}y \) or \( x = -\sqrt{3}y \). We will use \( x = \sqrt{3}y \) for our calculations.

Substituting \( x = \sqrt{3}y \) into the imaginary part equation:
\[ 3(\sqrt{3}y)^2y - y^3 = 8 \]
Calculating \( 3(\sqrt{3}y)^2y \):
\[ 3 \cdot 3y^2y = 9y^3 \]
Thus, we have:
\[ 9y^3 - y^3 = 8 \]
This simplifies to:
\[ 8y^3 = 8 \]
Dividing both sides by 8:
\[ y^3 = 1 \]
Taking the cube root:
\[ y = 1 \]

### Conclusion
The maximum value of \( y \) is:
\[ \boxed{1} \]