The number of complex numbers satisfying (1 + i)z = i|z|

Options

A. 0
B. 1
C. 2
D. infinite

Answer

To solve the equation \((1 + i)z = i|z|\) for the number of complex numbers \(z\) that satisfy it, we can follow these steps:

### Step 1: Express \(z\) in terms of real and imaginary parts
Let \(z = x + iy\), where \(x\) and \(y\) are real numbers.

### Step 2: Calculate the modulus of \(z\)
The modulus of \(z\) is given by:
\[ |z| = \sqrt{x^2 + y^2} \]

### Step 3: Substitute \(z\) and \(|z|\) into the equation
Substituting \(z\) and \(|z|\) into the equation, we have:
\[ (1 + i)(x + iy) = i\sqrt{x^2 + y^2} \]

### Step 4: Expand the left-hand side
Expanding the left-hand side:
\[ (1 + i)(x + iy) = x + iy + ix - y = (x - y) + i(x + y) \]
Thus, the equation becomes:
\[ (x - y) + i(x + y) = i\sqrt{x^2 + y^2} \]

### Step 5: Equate real and imaginary parts
For the two complex numbers to be equal, their real parts and imaginary parts must be equal. Therefore, we have:
1. Real part: \(x - y = 0\)
2. Imaginary part: \(x + y = \sqrt{x^2 + y^2}\)

### Step 6: Solve the equations
From the first equation \(x - y = 0\), we can conclude:
\[ x = y \]
Substituting \(y = x\) into the second equation:
\[ x + x = \sqrt{x^2 + x^2} \]
This simplifies to:
\[ 2x = \sqrt{2x^2} \]
Squaring both sides:
\[ 4x^2 = 2x^2 \]
This leads to:
\[ 2x^2 = 0 \implies x^2 = 0 \implies x = 0 \]
Since \(x = y\), we also have \(y = 0\).

### Step 7: Conclusion
Thus, the only solution for \(z\) is:
\[ z = 0 + 0i = 0 \]
Therefore, there is only **one complex number** that satisfies the given equation.

### Final Answer
The number of complex numbers satisfying \((1 + i)z = i|z|\) is **1**.

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