The modulus of `((1+2i)(3-4i))/((4+3i)(2-3i))`is

Options

A. `sqrt((5)/(13))`
B. `sqrt((13)/(5))`
C. 1
D. `-1`

Answer

To find the modulus of the expression \(\frac{(1+2i)(3-4i)}{(4+3i)(2-3i)}\), we can follow these steps:

### Step 1: Identify the components
Let:
- \( z_1 = (1 + 2i)(3 - 4i) \)
- \( z_2 = (4 + 3i)(2 - 3i) \)

### Step 2: Calculate the modulus of \( z_1 \)
The modulus of a complex number \( a + bi \) is given by \( |z| = \sqrt{a^2 + b^2} \).

First, we calculate \( z_1 \):
\[ z_1 = (1 + 2i)(3 - 4i) = 1 \cdot 3 + 1 \cdot (-4i) + 2i \cdot 3 + 2i \cdot (-4i) \]
\[ = 3 - 4i + 6i - 8(-1) = 3 - 4i + 6i + 8 = 11 + 2i \]

Now, calculate the modulus of \( z_1 \):
\[ |z_1| = |11 + 2i| = \sqrt{11^2 + 2^2} = \sqrt{121 + 4} = \sqrt{125} = 5\sqrt{5} \]

### Step 3: Calculate the modulus of \( z_2 \)
Now, we calculate \( z_2 \):
\[ z_2 = (4 + 3i)(2 - 3i) = 4 \cdot 2 + 4 \cdot (-3i) + 3i \cdot 2 + 3i \cdot (-3i) \]
\[ = 8 - 12i + 6i - 9(-1) = 8 - 12i + 6i + 9 = 17 - 6i \]

Now, calculate the modulus of \( z_2 \):
\[ |z_2| = |17 - 6i| = \sqrt{17^2 + (-6)^2} = \sqrt{289 + 36} = \sqrt{325} = 5\sqrt{13} \]

### Step 4: Calculate the modulus of the quotient
Now we find the modulus of the quotient:
\[ \left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|} = \frac{5\sqrt{5}}{5\sqrt{13}} = \frac{\sqrt{5}}{\sqrt{13}} = \sqrt{\frac{5}{13}} \]

### Final Answer
Thus, the modulus of \(\frac{(1+2i)(3-4i)}{(4+3i)(2-3i)}\) is:
\[ \sqrt{\frac{5}{13}} \]