If z=x+iy, where `i=sqrt(-1)`, then the equation `abs(((2z-i)/(z+1)))=m` represents a circle, then m can be

Options

A. `1/2`
B. 1
C. 2
D. `in(3,2sqrt(3))`

Answer

To solve the problem, we need to analyze the given equation involving complex numbers. Let's break it down step by step.

### Step 1: Substitute the expression for z
Given \( z = x + iy \), we substitute this into the equation:

\[ \left| \frac{2z - i}{z + 1} \right| = m \]

Substituting \( z \):

\[ \left| \frac{2(x + iy) - i}{(x + iy) + 1} \right| = m \]

### Step 2: Simplify the numerator and denominator
Now, simplify the numerator and denominator:

Numerator:
\[ 2(x + iy) - i = 2x + 2iy - i = 2x + i(2y - 1) \]

Denominator:
\[ (x + iy) + 1 = (x + 1) + iy \]

Thus, we have:

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

### Step 3: Cross multiply to eliminate the modulus
Cross multiplying gives us:

\[ |2x + i(2y - 1)| = m |(x + 1) + iy| \]

### Step 4: Calculate the moduli
Now we calculate the moduli:

For the left side:

\[ |2x + i(2y - 1)| = \sqrt{(2x)^2 + (2y - 1)^2} = \sqrt{4x^2 + (2y - 1)^2} \]

For the right side:

\[ |(x + 1) + iy| = \sqrt{(x + 1)^2 + y^2} \]

Thus, we have:

\[ \sqrt{4x^2 + (2y - 1)^2} = m \sqrt{(x + 1)^2 + y^2} \]

### Step 5: Square both sides
Squaring both sides to eliminate the square roots gives:

\[ 4x^2 + (2y - 1)^2 = m^2 \left((x + 1)^2 + y^2\right) \]

### Step 6: Expand both sides
Expanding both sides:

Left side:
\[ 4x^2 + (4y^2 - 4y + 1) = 4x^2 + 4y^2 - 4y + 1 \]

Right side:
\[ m^2 \left(x^2 + 2x + 1 + y^2\right) = m^2(x^2 + y^2 + 2x + 1) \]

### Step 7: Rearranging the equation
Now, we rearrange the equation:

\[ 4x^2 + 4y^2 - 4y + 1 = m^2(x^2 + y^2 + 2x + 1) \]

### Step 8: Collect like terms
Collecting like terms leads to:

\[ (4 - m^2)x^2 + (4 - m^2)y^2 - 2m^2x - 4y + (1 - m^2) = 0 \]

### Step 9: Identify conditions for a circle
For this equation to represent a circle, the coefficients of \( x^2 \) and \( y^2 \) must be equal, and the coefficient of \( xy \) must be zero. Thus, we need:

\[ 4 - m^2 = 0 \implies m^2 = 4 \implies m = \pm 2 \]

### Conclusion
For \( m^2 \) to be less than 4, the equation represents a circle. Therefore, the possible values for \( m \) that allow the equation to represent a circle are:

\[ m < 2 \text{ or } m > -2 \]

Thus, \( m \) can be \( 2 \) or \( -2 \) and any value between them.