If `|z-3+ i|=4`, then the locus of z is

Options

A. `x^(2) + y^(2)- 6= 0`
B. `x^(2) + y^(2) - 3x + y-6= 0`
C. `x^(2) + y^(2)- 6x - 2= 0`
D. `x^(2) + y^(2)- 6x + 2y -6= 0`

Answer

To find the locus of the complex number \( z \) given the equation \( |z - 3 + i| = 4 \), we can follow these steps:

### Step 1: Understand the given equation
The equation \( |z - 3 + i| = 4 \) represents the distance from the point \( 3 - i \) in the complex plane to the point \( z \). The modulus indicates that this distance is equal to 4.

### Step 2: Express \( z \) in terms of its real and imaginary parts
Let \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part. Then we can rewrite the equation as:
\[ | (x + iy) - (3 - i) | = 4 \]
This simplifies to:
\[ | (x - 3) + i(y + 1) | = 4 \]

### Step 3: Write the modulus in terms of \( x \) and \( y \)
The modulus of a complex number \( a + bi \) is given by \( \sqrt{a^2 + b^2} \). Therefore, we have:
\[ \sqrt{(x - 3)^2 + (y + 1)^2} = 4 \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (x - 3)^2 + (y + 1)^2 = 16 \]

### Step 5: Expand the equation
Expanding the left-hand side:
\[ (x^2 - 6x + 9) + (y^2 + 2y + 1) = 16 \]
This simplifies to:
\[ x^2 + y^2 - 6x + 2y + 10 = 16 \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ x^2 + y^2 - 6x + 2y - 6 = 0 \]

### Step 7: Identify the locus
The equation \( x^2 + y^2 - 6x + 2y - 6 = 0 \) represents a circle in the complex plane.

### Final Answer
Thus, the locus of \( z \) is given by the equation:
\[ x^2 + y^2 - 6x + 2y - 6 = 0 \]