The system of equations `|z+1-i|=sqrt2 and |z| = 3` has how many solutions?

Options

A. no solution
B. one solution
C. two solution
D. None of these

Answer

To solve the given system of equations \( |z + 1 - i| = \sqrt{2} \) and \( |z| = 3 \), we will follow these steps:

### Step 1: Define the Complex Number
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

### Step 2: Substitute \( z \) into the First Equation
Substituting \( z \) into the first equation gives:
\[ |z + 1 - i| = |(x + iy) + (1 - i)| = |(x + 1) + (y - 1)i| \]
This can be expressed as:
\[ \sqrt{(x + 1)^2 + (y - 1)^2} = \sqrt{2} \]

### Step 3: Square Both Sides
Squaring both sides results in:
\[ (x + 1)^2 + (y - 1)^2 = 2 \]

### Step 4: Expand the Equation
Expanding the left side gives:
\[ (x^2 + 2x + 1) + (y^2 - 2y + 1) = 2 \]
This simplifies to:
\[ x^2 + y^2 + 2x - 2y + 2 = 2 \]
Thus, we have:
\[ x^2 + y^2 + 2x - 2y = 0 \quad \text{(Equation 1)} \]

### Step 5: Use the Second Equation
From the second equation \( |z| = 3 \), we have:
\[ |z| = |x + iy| = \sqrt{x^2 + y^2} = 3 \]
Squaring both sides gives:
\[ x^2 + y^2 = 9 \quad \text{(Equation 2)} \]

### Step 6: Substitute Equation 2 into Equation 1
Substituting \( x^2 + y^2 = 9 \) into Equation 1:
\[ 9 + 2x - 2y = 0 \]
Rearranging gives:
\[ 2x - 2y = -9 \quad \Rightarrow \quad x - y = -\frac{9}{2} \quad \Rightarrow \quad x + y = \frac{9}{2} \quad \text{(Equation 3)} \]

### Step 7: Analyze the Equations
Now we have two equations:
1. \( x + y = \frac{9}{2} \) (Equation 3)
2. \( x^2 + y^2 = 9 \) (Equation 2)

### Step 8: Solve for \( y \) in terms of \( x \)
From Equation 3, we can express \( y \) in terms of \( x \):
\[ y = \frac{9}{2} - x \]

### Step 9: Substitute \( y \) into Equation 2
Substituting \( y \) into Equation 2:
\[ x^2 + \left(\frac{9}{2} - x\right)^2 = 9 \]
Expanding gives:
\[ x^2 + \left(\frac{81}{4} - 9x + x^2\right) = 9 \]
Combining terms results in:
\[ 2x^2 - 9x + \frac{81}{4} - 9 = 0 \]
This simplifies to:
\[ 2x^2 - 9x + \frac{45}{4} = 0 \]

### Step 10: Multiply through by 4 to eliminate the fraction
Multiplying the entire equation by 4 gives:
\[ 8x^2 - 36x + 45 = 0 \]

### Step 11: Use the Quadratic Formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{36 \pm \sqrt{(-36)^2 - 4 \cdot 8 \cdot 45}}{2 \cdot 8} \]
Calculating the discriminant:
\[ x = \frac{36 \pm \sqrt{1296 - 1440}}{16} = \frac{36 \pm \sqrt{-144}}{16} \]
Since the discriminant is negative, there are no real solutions for \( x \).

### Conclusion
Thus, the system of equations has **no solutions**.

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