If `a+ bi = (c + i)/(c-i), a, b , c in R`, show that `a^(2) + b^(2)=1 and (b)/(a)= (2c)/(c^(2)-1)`

Answer

To solve the problem, we need to show that if \( a + bi = \frac{c + i}{c - i} \), where \( a, b, c \in \mathbb{R} \), then \( a^2 + b^2 = 1 \) and \( \frac{b}{a} = \frac{2c}{c^2 - 1} \).

### Step 1: Find the conjugate of the expression
We start with the given expression:
\[ a + bi = \frac{c + i}{c - i} \]
The conjugate of \( a + bi \) is \( a - bi \). The conjugate of the right-hand side can be calculated as follows:
\[ \overline{\left(\frac{c + i}{c - i}\right)} = \frac{\overline{(c + i)}}{\overline{(c - i)}} = \frac{c - i}{c + i} \]

### Step 2: Multiply the expressions
Now, we multiply \( a + bi \) and \( a - bi \):
\[ (a + bi)(a - bi) = a^2 + b^2 \]
On the right-hand side, we have:
\[ \frac{c + i}{c - i} \cdot \frac{c - i}{c + i} = 1 \]
Thus, we can equate the two sides:
\[ a^2 + b^2 = 1 \]

### Step 3: Find the values of \( a \) and \( b \)
Next, we will find the individual values of \( a \) and \( b \). We can express \( a + bi \) and \( a - bi \) as:
\[ a + bi = \frac{c + i}{c - i} \]
\[ a - bi = \frac{c - i}{c + i} \]

### Step 4: Add the two equations
Adding the two equations:
\[ (a + bi) + (a - bi) = \frac{c + i}{c - i} + \frac{c - i}{c + i} \]
This simplifies to:
\[ 2a = \frac{(c + i)(c + i) + (c - i)(c - i)}{(c - i)(c + i)} \]
Calculating the numerator:
\[ (c + i)(c + i) = c^2 + 2ci - 1 \]
\[ (c - i)(c - i) = c^2 - 2ci - 1 \]
Adding these gives:
\[ 2c^2 - 2 = 2c^2 - 2 \]
Thus, we have:
\[ 2a = \frac{2(c^2 - 1)}{c^2 + 1} \]
So,
\[ a = \frac{c^2 - 1}{c^2 + 1} \]

### Step 5: Subtract the two equations
Now, subtracting the two equations:
\[ (a + bi) - (a - bi) = \frac{c + i}{c - i} - \frac{c - i}{c + i} \]
This simplifies to:
\[ 2bi = \frac{(c + i)(c + i) - (c - i)(c - i)}{(c - i)(c + i)} \]
Calculating the numerator:
\[ (c + i)(c + i) - (c - i)(c - i) = 2i(2c) \]
Thus, we have:
\[ 2bi = \frac{4ci}{c^2 + 1} \]
So,
\[ b = \frac{2c}{c^2 - 1} \]

### Conclusion
We have shown that:
1. \( a^2 + b^2 = 1 \)
2. \( \frac{b}{a} = \frac{2c}{c^2 - 1} \)