Consider the equation `az + b barz + c =0`, where a,b,c `in`Z
If `|a| = |b| and barac ne b barc`, then z has

Options

A. infnite solutions
B. no solutions
C. finite solutions
D. cannot say anything

Explanation

To solve the equation \( az + b \bar{z} + c = 0 \) under the conditions \( |a| = |b| \) and \( \bar{a}c \neq b \bar{c} \), we will follow these steps:

### Step 1: Write the original equation
We start with the equation:
\[ az + b \bar{z} + c = 0 \]

### Step 2: Take the conjugate of the entire equation
Taking the conjugate of both sides gives:
\[ \overline{az + b \bar{z} + c} = \overline{0} \]
This simplifies to:
\[ \overline{az} + \overline{b \bar{z}} + \overline{c} = 0 \]

### Step 3: Apply the properties of conjugates
Using the properties of conjugates, we can separate the terms:
\[ \bar{a} \bar{z} + \bar{b} z + \bar{c} = 0 \]

### Step 4: Express \( \bar{z} \) in terms of \( z \)
From the original equation \( az + b \bar{z} + c = 0 \), we can express \( \bar{z} \) as:
\[ b \bar{z} = -c - az \implies \bar{z} = \frac{-c - az}{b} \]

### Step 5: Substitute \( \bar{z} \) into the conjugate equation
Substituting \( \bar{z} \) into the conjugate equation gives:
\[ \bar{a} \left(\frac{-c - az}{b}\right) + \bar{b} z + \bar{c} = 0 \]

### Step 6: Multiply through by \( b \) to eliminate the denominator
Multiplying through by \( b \) results in:
\[ -\bar{a}c - a z + \bar{b} b z + \bar{c} b = 0 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ (-\bar{a}c + \bar{c}b) + (-a + |\bar{b}|^2) z = 0 \]

### Step 8: Factor out \( z \)
This can be rearranged to isolate \( z \):
\[ z \left(|b|^2 - |a|^2\right) = \bar{a}c - b \bar{c} \]

### Step 9: Analyze the conditions
Given that \( |a| = |b| \), we have \( |b|^2 - |a|^2 = 0 \). Thus, the equation simplifies to:
\[ 0 \cdot z = \bar{a}c - b \bar{c} \]

### Step 10: Conclusion based on the conditions
Since \( \bar{a}c \neq b \bar{c} \), the right side is not equal to zero. Therefore, we conclude that there is no solution for \( z \).

### Final Answer
Thus, \( z \) has **no solution**.

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