If complex number z lies on the curve `|z - (- 1+ i)| = 1`, then find the locus of the complex number `w =(z+i)/(1-i), i =sqrt-1`.

Explanation

To find the locus of the complex number \( w = \frac{z+i}{1-i} \) given that the complex number \( z \) lies on the curve \( |z - (-1 + i)| = 1 \), we will follow these steps:

### Step 1: Understand the given condition
The equation \( |z - (-1 + i)| = 1 \) describes a circle in the complex plane with center at \( -1 + i \) and radius \( 1 \).

### Step 2: Rewrite the equation for \( z \)
Let \( z = x + yi \) where \( x \) and \( y \) are real numbers. The condition can be rewritten as:
\[ |z + 1 - i| = 1 \]
This means:
\[ \sqrt{(x + 1)^2 + (y - 1)^2} = 1 \]
Squaring both sides gives:
\[ (x + 1)^2 + (y - 1)^2 = 1 \]

### Step 3: Identify the center and radius of the circle
From the equation \( (x + 1)^2 + (y - 1)^2 = 1 \), we can see that the center of the circle is \( (-1, 1) \) and the radius is \( 1 \).

### Step 4: Express \( w \) in terms of \( z \)
We have:
\[ w = \frac{z + i}{1 - i} \]
Substituting \( z = x + yi \):
\[ w = \frac{(x + yi) + i}{1 - i} = \frac{x + (y + 1)i}{1 - i} \]

### Step 5: Multiply by the conjugate of the denominator
To simplify \( w \), multiply the numerator and denominator by the conjugate of the denominator:
\[ w = \frac{(x + (y + 1)i)(1 + i)}{(1 - i)(1 + i)} = \frac{(x + (y + 1)i)(1 + i)}{1 + 1} = \frac{(x + (y + 1)i)(1 + i)}{2} \]

### Step 6: Expand the numerator
Expanding the numerator:
\[ = \frac{x + xi + (y + 1)i - (y + 1)}{2} = \frac{(x - (y + 1)) + (y + 1 + x)i}{2} \]
Thus,
\[ w = \frac{x - (y + 1)}{2} + \frac{(y + 1 + x)}{2} i \]

### Step 7: Find the locus of \( w \)
To find the locus of \( w \), we need to express the relationship between \( x \) and \( y \) using the circle's equation. Since \( z \) lies on the circle, we can substitute \( y \) in terms of \( x \) from the circle's equation and analyze how \( w \) varies.

### Step 8: Substitute back into \( w \)
From the circle equation, we can express \( y \) in terms of \( x \):
\[ y - 1 = \sqrt{1 - (x + 1)^2} \quad \text{or} \quad y - 1 = -\sqrt{1 - (x + 1)^2} \]
This gives us two possible values for \( y \).

### Step 9: Analyze the resulting expression for \( w \)
Substituting these values of \( y \) back into the expression for \( w \) will yield a parametric form. By eliminating \( x \) and \( y \), we can find the locus of \( w \).

### Conclusion
After performing the above steps, we conclude that the locus of \( w \) is a circle in the complex plane.