Let `A(2,0)` and `B(z)` are two points on the circle `|z|=2`. `M(z')` is the point on `AB`. If the point `barz'` lies on the median of the triangle `OAB` where `O` is origin, then `arg(z')` is

Options

A. `tan^(-1)((sqrt(15))/(5))`
B. `tan^(-1)(sqrt(15))`
C. `tan^(-1)((5)/(sqrt(15)))`
D. `(pi)/(2)`

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript while ensuring clarity and completeness in each step.

### Step-by-Step Solution:

1. **Identify Points and Circle**:
Let the points be \( A(2, 0) \) and \( B(z) \), where \( |z| = 2 \). This means that point \( B \) lies on a circle of radius 2 centered at the origin.

2. **Midpoint of Segment AB**:
The point \( M(z') \) is the midpoint of segment \( AB \). The coordinates of \( M(z') \) can be calculated as:
\[ z' = \frac{A + B}{2} = \frac{2 + z}{2} \]

3. **Condition for \( \bar{z}' \)**:
We are given that the point \( \bar{z}' \) lies on the median of triangle \( OAB \), where \( O \) is the origin. The median from \( O \) to side \( AB \) can be expressed in terms of the coordinates of points \( A \) and \( B \).

4. **Collinearity Condition**:
The points \( O(0, 0) \), \( A(2, 0) \), and \( B(z) \) are collinear if the determinant of the matrix formed by these points is zero:
\[ \begin{vmatrix} 0 & 0 & 1 \\ 2 & 0 & 1 \\ z & \bar{z} & 1 \end{vmatrix} = 0 \]

5. **Calculate the Determinant**:
Expanding the determinant gives:
\[ 0(0 - \bar{z}) - 2(0 - z) + 1(0 - 2\bar{z}) = 0 \]
Simplifying this, we get:
\[ -2z + 2\bar{z} = 0 \implies z = \bar{z} \]

6. **Real Part Condition**:
Since \( z = \bar{z} \), it implies that \( z \) is a real number. However, since \( |z| = 2 \), we can express \( z \) in terms of its real part:
\[ z + \bar{z} = 2 \implies \text{Re}(z) = 1 \]

7. **Finding the Imaginary Part**:
Since \( |z| = 2 \):
\[ \text{Re}(z)^2 + \text{Im}(z)^2 = 4 \implies \left(\frac{1}{2}\right)^2 + \text{Im}(z)^2 = 4 \]
This leads to:
\[ \frac{1}{4} + \text{Im}(z)^2 = 4 \implies \text{Im}(z)^2 = 4 - \frac{1}{4} = \frac{15}{4} \implies \text{Im}(z) = \pm \frac{\sqrt{15}}{2} \]

8. **Final Form of \( z \)**:
Thus, we have:
\[ z = \frac{1}{2} + i\frac{\sqrt{15}}{2} \]

9. **Finding the Argument**:
The argument of \( z \) can be calculated using:
\[ \arg(z) = \tan^{-1}\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right) = \tan^{-1}\left(\frac{\frac{\sqrt{15}}{2}}{\frac{1}{2}}\right) = \tan^{-1}(\sqrt{15}) \]

### Final Answer:
\[ \arg(z) = \tan^{-1}(\sqrt{15}) \]