`(2-i)/(5-2i)`
If the expression above is rewritten in the form a+bi, where a and b are real numbers, what is the value of `-b` written as a fraction? (Note: `i^(2)=-1`)

Answer

To solve the expression \((2-i)/(5-2i)\) and rewrite it in the form \(a+bi\), where \(a\) and \(b\) are real numbers, we will follow these steps:

### Step 1: Multiply by the Conjugate
To eliminate the imaginary part in the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(5 - 2i\) is \(5 + 2i\).

\[ \frac{2 - i}{5 - 2i} \cdot \frac{5 + 2i}{5 + 2i} \]

### Step 2: Expand the Numerator
Now, we will expand the numerator:

\[ (2 - i)(5 + 2i) = 2 \cdot 5 + 2 \cdot 2i - i \cdot 5 - i \cdot 2i \]
\[ = 10 + 4i - 5i - 2i^2 \]

Since \(i^2 = -1\), we can substitute that in:

\[ = 10 + 4i - 5i + 2 = 12 - i \]

### Step 3: Expand the Denominator
Next, we will expand the denominator:

\[ (5 - 2i)(5 + 2i) = 5^2 - (2i)^2 = 25 - 4i^2 \]
\[ = 25 - 4(-1) = 25 + 4 = 29 \]

### Step 4: Combine the Results
Now we can combine the results from the numerator and denominator:

\[ \frac{12 - i}{29} = \frac{12}{29} - \frac{1}{29}i \]

### Step 5: Identify \(a\) and \(b\)
From the expression \(\frac{12}{29} - \frac{1}{29}i\), we can identify:
- \(a = \frac{12}{29}\)
- \(b = -\frac{1}{29}\)

### Step 6: Find \(-b\)
We need to find \(-b\):

\[ -b = -\left(-\frac{1}{29}\right) = \frac{1}{29} \]

### Final Answer
Thus, the value of \(-b\) written as a fraction is:

\[ \frac{1}{29} \]

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