Find the modulus of ` (2 -3i)/( 4+i)`

Answer

To find the modulus of the complex number \(\frac{2 - 3i}{4 + i}\), we can follow these steps:

### Step 1: Find the modulus of the numerator and the denominator separately.

The modulus of a complex number \(a + bi\) is given by \(\sqrt{a^2 + b^2}\).

**Numerator:**
For \(2 - 3i\):
- \(a = 2\)
- \(b = -3\)

\[ \text{Modulus of } (2 - 3i) = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \]

**Denominator:**
For \(4 + i\):
- \(a = 4\)
- \(b = 1\)

\[ \text{Modulus of } (4 + i) = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17} \]

### Step 2: Use the property of moduli of complex numbers.

The modulus of the quotient of two complex numbers is the quotient of their moduli:

\[ \left| \frac{2 - 3i}{4 + i} \right| = \frac{|2 - 3i|}{|4 + i|} = \frac{\sqrt{13}}{\sqrt{17}} \]

### Step 3: Rationalize the denominator.

To rationalize the denominator, multiply the numerator and the denominator by \(\sqrt{17}\):

\[ \frac{\sqrt{13}}{\sqrt{17}} \cdot \frac{\sqrt{17}}{\sqrt{17}} = \frac{\sqrt{13} \cdot \sqrt{17}}{17} = \frac{\sqrt{221}}{17} \]

### Final Answer:

The modulus of \(\frac{2 - 3i}{4 + i}\) is:

\[ \frac{\sqrt{221}}{17} \]

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