If `(1+2i) (2+3i)(3+4i)=x+iy,x,y in R ` then `x^(2)+y^(2)` is

Options

A. 1450
B. 1625
C. 1575
D. 1725

Answer

To solve the problem, we need to calculate the product of the complex numbers \((1 + 2i)\), \((2 + 3i)\), and \((3 + 4i)\) and find \(x^2 + y^2\) where \(x + iy\) is the result of the multiplication.

### Step-by-Step Solution:

1. **Multiply the first two complex numbers:**
\[ (2 + 3i)(3 + 4i) \]
Using the distributive property (FOIL method):
\[ = 2 \cdot 3 + 2 \cdot 4i + 3i \cdot 3 + 3i \cdot 4i \]
\[ = 6 + 8i + 9i + 12i^2 \]
Since \(i^2 = -1\):
\[ = 6 + 17i - 12 = -6 + 17i \]

2. **Now multiply the result with the third complex number:**
\[ (1 + 2i)(-6 + 17i) \]
Again using the distributive property:
\[ = 1 \cdot -6 + 1 \cdot 17i + 2i \cdot -6 + 2i \cdot 17i \]
\[ = -6 + 17i - 12i + 34i^2 \]
Replacing \(i^2\) with \(-1\):
\[ = -6 + 5i - 34 = -40 + 5i \]

3. **Identify \(x\) and \(y\):**
From the expression \(-40 + 5i\), we can see:
\[ x = -40 \quad \text{and} \quad y = 5 \]

4. **Calculate \(x^2 + y^2\):**
\[ x^2 + y^2 = (-40)^2 + 5^2 \]
\[ = 1600 + 25 = 1625 \]

### Final Answer:
Thus, the value of \(x^2 + y^2\) is \(1625\).

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