If a +ib = `((x+i)^(2))/(2x-1)` then `a^(2)+b^(2)` is equal to

Options

A. `((x+1)^(4))/(4x^(2)-1)`
B. `((x^2+1)^(2))/(2x-1)^2`
C. `((x+1)^(4))/(4x^(2)+1)`
D. `((x+1)^(2))/(4x^(2)-1)`

Answer

To solve the problem where \( a + ib = \frac{(x + i)^2}{2x - 1} \) and we need to find \( a^2 + b^2 \), we can follow these steps:

### Step 1: Expand the numerator
We start with the expression:
\[ a + ib = \frac{(x + i)^2}{2x - 1} \]
Now, we expand \( (x + i)^2 \):
\[ (x + i)^2 = x^2 + 2xi + i^2 \]
Since \( i^2 = -1 \), we can write:
\[ (x + i)^2 = x^2 + 2xi - 1 = x^2 - 1 + 2xi \]

### Step 2: Substitute back into the equation
Now substitute the expanded form back into the equation:
\[ a + ib = \frac{x^2 - 1 + 2xi}{2x - 1} \]

### Step 3: Separate real and imaginary parts
We can separate the real and imaginary parts:
\[ a = \frac{x^2 - 1}{2x - 1}, \quad b = \frac{2x}{2x - 1} \]

### Step 4: Find \( a^2 + b^2 \)
Now we need to compute \( a^2 + b^2 \):
\[ a^2 + b^2 = \left(\frac{x^2 - 1}{2x - 1}\right)^2 + \left(\frac{2x}{2x - 1}\right)^2 \]
The common denominator is \( (2x - 1)^2 \), so we can write:
\[ a^2 + b^2 = \frac{(x^2 - 1)^2 + (2x)^2}{(2x - 1)^2} \]

### Step 5: Expand the numerator
Now we expand the numerator:
\[ (x^2 - 1)^2 = x^4 - 2x^2 + 1 \]
\[ (2x)^2 = 4x^2 \]
Thus, we have:
\[ a^2 + b^2 = \frac{x^4 - 2x^2 + 1 + 4x^2}{(2x - 1)^2} \]
Combine like terms:
\[ = \frac{x^4 + 2x^2 + 1}{(2x - 1)^2} \]

### Step 6: Recognize the perfect square
Notice that the numerator can be factored:
\[ x^4 + 2x^2 + 1 = (x^2 + 1)^2 \]
Thus, we can write:
\[ a^2 + b^2 = \frac{(x^2 + 1)^2}{(2x - 1)^2} \]

### Final Result
Therefore, the final expression for \( a^2 + b^2 \) is:
\[ a^2 + b^2 = \left(\frac{x^2 + 1}{2x - 1}\right)^2 \]

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