If the points in the complex plane satisfy the equations `log_(5)(|z|+3)-log_(sqrt(5))(|z-1|)=1` and arg `(z-1)=pi/4` are of the form `A_(1)+iB_(1)`, then the value of `A_(1)+B_(1)`, is

Options

A. `2sqrt(2)`
B. `sqrt(2)`
C. `4sqrt(2)`
D. 0

Answer

To solve the given problem, we will break it down into manageable steps.

### Step 1: Rewrite the logarithmic equation
We start with the equation:
\[ \log_{5}(|z| + 3) - \log_{\sqrt{5}}(|z - 1|) = 1 \]
We can convert the logarithm to the same base. Recall that:
\[ \log_{\sqrt{5}}(x) = \frac{2}{1} \log_{5}(x) = 2 \log_{5}(x) \]
Thus, we can rewrite the equation as:
\[ \log_{5}(|z| + 3) - 2 \log_{5}(|z - 1|) = 1 \]

### Step 2: Combine the logarithmic terms
Using the properties of logarithms, we can combine the terms:
\[ \log_{5}\left(\frac{|z| + 3}{|z - 1|^2}\right) = 1 \]
This implies:
\[ \frac{|z| + 3}{|z - 1|^2} = 5 \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ |z| + 3 = 5 |z - 1|^2 \]

### Step 4: Expressing in terms of |z|
Let \( |z| = r \). Then we have:
\[ r + 3 = 5 |z - 1|^2 \]
We need to express \( |z - 1| \) in terms of \( r \). If we let \( z = x + iy \), then:
\[ |z - 1| = |(x - 1) + iy| = \sqrt{(x - 1)^2 + y^2} \]
Thus, we can write:
\[ r + 3 = 5((x - 1)^2 + y^2) \]

### Step 5: Using the second condition
The second condition given is:
\[ \arg(z - 1) = \frac{\pi}{4} \]
This means that:
\[ \frac{y}{x - 1} = 1 \quad \Rightarrow \quad y = x - 1 \]

### Step 6: Substitute y in terms of x
Substituting \( y = x - 1 \) into the equation for \( r \):
\[ r = \sqrt{x^2 + (x - 1)^2} = \sqrt{x^2 + (x^2 - 2x + 1)} = \sqrt{2x^2 - 2x + 1} \]
Now substituting this into the equation:
\[ \sqrt{2x^2 - 2x + 1} + 3 = 5((x - 1)^2 + (x - 1)^2) \]
This simplifies to:
\[ \sqrt{2x^2 - 2x + 1} + 3 = 10(x - 1)^2 \]

### Step 7: Solve for x
Squaring both sides and solving for \( x \):
1. Isolate the square root.
2. Square both sides.
3. Simplify and solve the resulting quadratic equation.

### Step 8: Find values of A and B
Once we have \( x \), we can find \( y \) using \( y = x - 1 \). The values \( A_1 \) and \( B_1 \) correspond to \( x \) and \( y \) respectively.

### Step 9: Calculate \( A_1 + B_1 \)
Finally, compute:
\[ A_1 + B_1 = x + (x - 1) = 2x - 1 \]