If z = x +iy satisfies |z+1|=1 then

Options

A. (a) x=0
B. (b) `(x-1)^(2)+y^(2)=1 `
C. (c) `y=0`
D. (d) `(x+1)^(2)+y^(2)=1 `

Answer

To solve the problem where \( z = x + iy \) satisfies \( |z + 1| = 1 \), we can follow these steps:

### Step 1: Substitute \( z \)
We start with the given expression:
\[ z + 1 = (x + iy) + 1 = (x + 1) + iy \]

### Step 2: Write the modulus
The modulus of a complex number \( a + bi \) is given by:
\[ |a + bi| = \sqrt{a^2 + b^2} \]
Applying this to our expression, we have:
\[ |z + 1| = |(x + 1) + iy| = \sqrt{(x + 1)^2 + y^2} \]

### Step 3: Set up the equation
According to the problem, we know:
\[ |z + 1| = 1 \]
Thus, we can set up the equation:
\[ \sqrt{(x + 1)^2 + y^2} = 1 \]

### Step 4: Square both sides
To eliminate the square root, we square both sides:
\[ ((x + 1)^2 + y^2) = 1^2 \]
This simplifies to:
\[ (x + 1)^2 + y^2 = 1 \]

### Conclusion
The final equation we derived is:
\[ (x + 1)^2 + y^2 = 1 \]
This matches option \( d \) from the given choices.

### Final Answer
Thus, the correct option is:
\[ \text{Option D: } (x + 1)^2 + y^2 = 1 \]

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