If `i=sqrt(-1), and ((7+5i))/((-2-6i))=a+bi`, where a and b are real numbers, then what is the value of `|a+b|`?

Answer

To solve the problem, we need to simplify the expression \(\frac{7 + 5i}{-2 - 6i}\) and express it in the form \(a + bi\), where \(a\) and \(b\) are real numbers. Then, we will find the value of \(|a + b|\).

### Step 1: Rewrite the denominator
The denominator is \(-2 - 6i\). We can factor out \(-1\) from the denominator:
\[ \frac{7 + 5i}{-2 - 6i} = \frac{7 + 5i}{-1(2 + 6i)} = -\frac{7 + 5i}{2 + 6i} \]

### Step 2: Multiply by the conjugate
To simplify the expression, we multiply the numerator and the denominator by the conjugate of the denominator, which is \(2 - 6i\):
\[ -\frac{(7 + 5i)(2 - 6i)}{(2 + 6i)(2 - 6i)} \]

### Step 3: Simplify the denominator
Using the identity \( (a + bi)(a - bi) = a^2 + b^2 \):
\[ (2 + 6i)(2 - 6i) = 2^2 + 6^2 = 4 + 36 = 40 \]

### Step 4: Simplify the numerator
Now, we will expand the numerator:
\[ (7 + 5i)(2 - 6i) = 7 \cdot 2 + 7 \cdot (-6i) + 5i \cdot 2 + 5i \cdot (-6i) \]
Calculating each term:
\[ = 14 - 42i + 10i - 30i^2 \]
Since \(i^2 = -1\), we substitute:
\[ = 14 - 42i + 10i + 30 = 44 - 32i \]

### Step 5: Combine results
Now, substituting back into our expression:
\[ -\frac{44 - 32i}{40} = -\frac{44}{40} + \frac{32i}{40} \]
This simplifies to:
\[ -\frac{11}{10} + \frac{4}{5}i \]

### Step 6: Identify \(a\) and \(b\)
From the expression \(-\frac{11}{10} + \frac{4}{5}i\), we identify:
\[ a = -\frac{11}{10}, \quad b = \frac{4}{5} \]

### Step 7: Calculate \(a + b\)
Now, we calculate \(a + b\):
\[ a + b = -\frac{11}{10} + \frac{4}{5} \]
To add these, we convert \(\frac{4}{5}\) to have a common denominator of 10:
\[ \frac{4}{5} = \frac{8}{10} \]
Thus,
\[ a + b = -\frac{11}{10} + \frac{8}{10} = -\frac{3}{10} \]

### Step 8: Find \(|a + b|\)
Finally, we find the absolute value:
\[ |a + b| = \left| -\frac{3}{10} \right| = \frac{3}{10} \]

### Final Answer
The value of \(|a + b|\) is \(\frac{3}{10}\) or \(0.3\).

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