Find the modulus of the following using the property of modulus
`((2-3i)(4+5i))/((1-4i)(2-i))`

Answer

To find the modulus of the complex number given by the expression \(\frac{(2-3i)(4+5i)}{(1-4i)(2-i)}\), we can use the property of modulus, which states that the modulus of a quotient of complex numbers is equal to the quotient of their moduli:

\[ |z| = \frac{|a|}{|b|} \]

where \(z = \frac{a}{b}\).

### Step 1: Calculate the modulus of the numerator \((2-3i)(4+5i)\)

First, we calculate the modulus of each factor in the numerator.

1. For \(2 - 3i\):
\[ |2 - 3i| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \]

2. For \(4 + 5i\):
\[ |4 + 5i| = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \]

Now, the modulus of the numerator is:
\[ |(2-3i)(4+5i)| = |2-3i| \cdot |4+5i| = \sqrt{13} \cdot \sqrt{41} = \sqrt{13 \cdot 41} = \sqrt{533} \]

### Step 2: Calculate the modulus of the denominator \((1-4i)(2-i)\)

Next, we calculate the modulus of each factor in the denominator.

1. For \(1 - 4i\):
\[ |1 - 4i| = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \]

2. For \(2 - i\):
\[ |2 - i| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

Now, the modulus of the denominator is:
\[ |(1-4i)(2-i)| = |1-4i| \cdot |2-i| = \sqrt{17} \cdot \sqrt{5} = \sqrt{17 \cdot 5} = \sqrt{85} \]

### Step 3: Calculate the modulus of the entire expression

Now we can find the modulus of the entire expression:
\[ \left|\frac{(2-3i)(4+5i)}{(1-4i)(2-i)}\right| = \frac{|(2-3i)(4+5i)|}{|(1-4i)(2-i)|} = \frac{\sqrt{533}}{\sqrt{85}} = \sqrt{\frac{533}{85}} \]

### Final Result

Thus, the modulus of the given complex number is:
\[ \sqrt{\frac{533}{85}} \]