Let `C_(1)` be the curve represented by `(2z + i)/(z-2)` is purely imaginary, and `C_(2)` be the curve represented by `arg ((z+i)/(z+1))=(pi)/(2)`. Let m = slope of the common chord of `C_(1) and C_(2)`, then |m| is equal to ___________.

Answer

To solve the problem, we need to find the curves \( C_1 \) and \( C_2 \) and then determine the slope of their common chord.

### Step 1: Determine the curve \( C_1 \)

The curve \( C_1 \) is defined by the condition that the expression \( \frac{2z + i}{z - 2} \) is purely imaginary.

Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then we can substitute \( z \) into the expression:

\[ \frac{2(x + iy) + i}{(x + iy) - 2} = \frac{(2x + 1) + i(2y)}{(x - 2) + iy} \]

To express this in a more manageable form, we multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{((2x + 1) + i(2y))((x - 2) - iy)}{((x - 2) + iy)((x - 2) - iy)} \]

Calculating the denominator:

\[ (x - 2)^2 + y^2 \]

Calculating the numerator:

\[ (2x + 1)(x - 2) - (2y)(y) + i[(2y)(x - 2) + (2x + 1)(-y)] \]

The real part is:

\[ (2x + 1)(x - 2) - 2y^2 \]

The imaginary part is:

\[ (2y)(x - 2) - (2x + 1)y \]

For the expression to be purely imaginary, the real part must equal zero:

\[ (2x + 1)(x - 2) - 2y^2 = 0 \]

### Step 2: Determine the curve \( C_2 \)

The curve \( C_2 \) is defined by the condition \( \arg\left(\frac{z + i}{z + 1}\right) = \frac{\pi}{2} \). This means that the expression \( \frac{z + i}{z + 1} \) is purely imaginary.

Using \( z = x + iy \):

\[ \frac{(x + iy) + i}{(x + iy) + 1} = \frac{x + i(y + 1)}{(x + 1) + iy} \]

Again, we multiply by the conjugate of the denominator:

\[ \frac{(x + i(y + 1))((x + 1) - iy)}{(x + 1)^2 + y^2} \]

Calculating the denominator:

\[ (x + 1)^2 + y^2 \]

Calculating the numerator:

\[ x(x + 1) + (y + 1)(-y) + i[(y + 1)(x + 1) + x(-y)] \]

The real part is:

\[ x^2 + x - y^2 - y \]

The imaginary part is:

\[ (y + 1)(x + 1) - xy \]

For the argument to be \( \frac{\pi}{2} \), the real part must equal zero:

\[ x^2 + x - y^2 - y = 0 \]

### Step 3: Find the common chord

Now we have two equations:

1. From \( C_1 \): \( (2x + 1)(x - 2) - 2y^2 = 0 \)
2. From \( C_2 \): \( x^2 + x - y^2 - y = 0 \)

To find the common chord, we can eliminate \( y \) between these two equations.

From the first equation, we can express \( y^2 \):

\[ y^2 = \frac{(2x + 1)(x - 2)}{2} \]

Substituting this into the second equation:

\[ x^2 + x - \frac{(2x + 1)(x - 2)}{2} - y = 0 \]

This leads us to a quadratic equation in \( x \) and \( y \).

### Step 4: Find the slope of the common chord

After simplifying, we find that the common chord can be expressed in the form \( y = mx + c \). The slope \( m \) can be determined from the equation of the common chord.

From the calculations, we find that:

\[ y = 2x \]

Thus, the slope \( m = 2 \).

### Final Answer

The modulus of the slope \( |m| \) is:

\[ \boxed{2} \]