Let Z be a complex number satisfying the relation `Z^(3)+(4(barZ)^(2))/(|Z|)=0`. If the least possible argument of Z is `-kpi`, then k is equal to (here, `argZ in (-pi, pi]`)

Answer

To solve the problem, we start with the equation given:

\[ Z^3 + \frac{4(\bar{Z})^2}{|Z|} = 0 \]

1. **Express Z in Polar Form**:
Let \( Z = re^{i\theta} \), where \( r = |Z| \) and \( \theta = \arg(Z) \). The conjugate \( \bar{Z} \) can be expressed as \( \bar{Z} = re^{-i\theta} \).

2. **Substitute Z and \(\bar{Z}\) into the Equation**:
Substitute \( Z \) and \( \bar{Z} \) into the equation:

\[ (re^{i\theta})^3 + \frac{4(re^{-i\theta})^2}{r} = 0 \]

This simplifies to:

\[ r^3 e^{3i\theta} + \frac{4r^2 e^{-2i\theta}}{r} = 0 \]

Which further simplifies to:

\[ r^3 e^{3i\theta} + 4r e^{-2i\theta} = 0 \]

3. **Rearrange the Equation**:
Rearranging gives:

\[ r^3 e^{3i\theta} = -4r e^{-2i\theta} \]

Dividing both sides by \( r \) (assuming \( r \neq 0 \)):

\[ r^2 e^{3i\theta} = -4 e^{-2i\theta} \]

4. **Multiply Both Sides by \( e^{2i\theta} \)**:
This gives:

\[ r^2 e^{5i\theta} = -4 \]

5. **Take the Argument of Both Sides**:
The argument of the left side is \( 5\theta \) and the argument of the right side is \( \pi \) (since \(-4\) lies on the negative real axis):

\[ 5\theta = \pi + 2k\pi \quad (k \in \mathbb{Z}) \]

Therefore:

\[ \theta = \frac{\pi}{5} + \frac{2k\pi}{5} \]

6. **Find the Possible Values of \(\theta\)**:
For \( k = 0 \):
\[ \theta = \frac{\pi}{5} \]

For \( k = -1 \):
\[ \theta = -\frac{3\pi}{5} \]

For \( k = 1 \):
\[ \theta = \frac{5\pi}{5} = \pi \]

For \( k = 2 \):
\[ \theta = \frac{7\pi}{5} \quad \text{(not in } (-\pi, \pi] \text{)} \]

Thus, the possible values of \( \theta \) are:
\[ \frac{\pi}{5}, -\frac{3\pi}{5}, \pi \]

7. **Identify the Least Possible Argument**:
The least possible argument in the range \((- \pi, \pi]\) is \(-\frac{3\pi}{5}\).

8. **Determine the Value of k**:
Given that the least possible argument of \( Z \) is \(-k\pi\), we have:

\[ -k\pi = -\frac{3\pi}{5} \implies k = \frac{3}{5} \]

Thus, the value of \( k \) is:

\[ \boxed{\frac{3}{5}} \]