The equation `z^(2)-i|z-1|^(2)=0,` where `i=sqrt(-1),`has.

Options

A. no real root
B. no purely imaginary root
C. all roots inside `|z|=1`
D. atleast two roots

Answer

To solve the equation \( z^2 - i|z - 1|^2 = 0 \), where \( i = \sqrt{-1} \), we will follow these steps:

### Step 1: Substitute \( z \) with \( x + iy \)
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

### Step 2: Rewrite the equation
The equation becomes:
\[ (x + iy)^2 - i|x + iy - 1|^2 = 0 \]

### Step 3: Expand \( (x + iy)^2 \)
Calculating \( (x + iy)^2 \):
\[ (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + (2xy)i \]

### Step 4: Calculate \( |z - 1|^2 \)
Next, we need to calculate \( |z - 1|^2 \):
\[ |z - 1|^2 = |(x - 1) + iy|^2 = (x - 1)^2 + y^2 \]

### Step 5: Substitute back into the equation
Now substituting back into the equation:
\[ (x^2 - y^2) + (2xy)i - i((x - 1)^2 + y^2) = 0 \]

### Step 6: Separate real and imaginary parts
This gives us:
\[ (x^2 - y^2) + (2xy - (x - 1)^2 - y^2)i = 0 \]
From this, we can separate the real and imaginary parts:
1. Real part: \( x^2 - y^2 = 0 \)
2. Imaginary part: \( 2xy - (x - 1)^2 - y^2 = 0 \)

### Step 7: Solve the real part
From the real part \( x^2 - y^2 = 0 \), we get:
\[ x^2 = y^2 \implies x = y \text{ or } x = -y \]

### Step 8: Case 1: \( x = y \)
Substituting \( x = y \) into the imaginary part:
\[ 2y^2 - (y - 1)^2 - y^2 = 0 \]
Expanding:
\[ 2y^2 - (y^2 - 2y + 1) - y^2 = 0 \]
Simplifying:
\[ 2y^2 - y^2 + 2y - 1 = 0 \implies y^2 + 2y - 1 = 0 \]
Using the quadratic formula:
\[ y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2} \]
Thus, \( y = -1 + \sqrt{2} \) or \( y = -1 - \sqrt{2} \). Correspondingly, \( x = -1 + \sqrt{2} \) or \( x = -1 - \sqrt{2} \).

### Step 9: Case 2: \( x = -y \)
Substituting \( x = -y \) into the imaginary part:
\[ 2(-y)y - ((-y) - 1)^2 - y^2 = 0 \]
This simplifies to:
\[ -2y^2 - (1 - 2y + y^2) - y^2 = 0 \implies -2y^2 - 1 + 2y - 2y^2 = 0 \]
This leads to:
\[ -4y^2 + 2y - 1 = 0 \]
Using the quadratic formula:
\[ y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot (-4) \cdot (-1)}}{2 \cdot (-4)} = \frac{-2 \pm \sqrt{4 - 16}}{-8} = \frac{-2 \pm \sqrt{-12}}{-8} \]
This gives complex solutions, indicating no purely imaginary roots.

### Step 10: Conclusion
Thus, we find that:
1. There are no real roots.
2. There are no purely imaginary roots.
3. The roots are within the unit circle \( |z| < 1 \).

### Final Answer
The equation has:
- No real roots
- No purely imaginary roots
- All roots inside \( |z| = 1 \)