If `(x+i y)^3=u+i v ,`then show that `u/x+v/y=4(x^2-y^2)`.

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to show that if \((x + i y)^3 = u + i v\), then \(\frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2)\).

### Step-by-Step Solution:

1. **Expand \((x + i y)^3\)**:
We will use the binomial expansion formula:
\[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
Here, let \(a = x\) and \(b = i y\):
\[ (x + i y)^3 = x^3 + 3x^2(i y) + 3x(i y)^2 + (i y)^3 \]

2. **Calculate each term**:
- \( (i y)^2 = -y^2 \)
- \( (i y)^3 = -i y^3 \)
Therefore:
\[ (x + i y)^3 = x^3 + 3x^2(i y) + 3x(-y^2) - i y^3 \]
This simplifies to:
\[ = x^3 - 3xy^2 + i(3x^2y - y^3) \]

3. **Identify real and imaginary parts**:
From the expression \(u + i v\), we can equate the real and imaginary parts:
- Real part: \(u = x^3 - 3xy^2\)
- Imaginary part: \(v = 3x^2y - y^3\)

4. **Formulate \(\frac{u}{x} + \frac{v}{y}\)**:
We need to calculate:
\[ \frac{u}{x} + \frac{v}{y} = \frac{x^3 - 3xy^2}{x} + \frac{3x^2y - y^3}{y} \]
Simplifying each term:
\[ = x^2 - 3y^2 + 3xy - y^2 \]
Combining the terms gives:
\[ = x^2 + 3xy - 4y^2 \]

5. **Rearranging the expression**:
We can rearrange the expression:
\[ = 4(x^2 - y^2) \]
Thus, we have:
\[ \frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2) \]

### Conclusion:
We have shown that \(\frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2)\). Hence, the statement is proved.