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We are given the function \( f(z) = (x^2 - 2xy + ay^2) + i(bx^2 - y^2 + 2xy) \), and we need to find constants \(a\) and \(b\) such that \( f(z) \) is analytic.

### Step 1: Understanding the function
The function \( f(z) \) is expressed in terms of \( x \) and \( y \), where:
- \( z = x + iy \) (so \( x \) and \( y \) are the real and imaginary parts of \( z \)).
- The real part of \( f(z) \) is \( u(x, y) = x^2 - 2xy + ay^2 \).
- The imaginary part of \( f(z) \) is \( v(x, y) = bx^2 - y^2 + 2xy \).

### Step 2: Applying the Cauchy-Riemann equations
For the function to be analytic, it must satisfy the Cauchy-Riemann equations:
1. \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \)
2. \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \)

#### First Cauchy-Riemann equation:
We start by calculating the partial derivatives.

- \( u(x, y) = x^2 - 2xy + ay^2 \)
  \[
  \frac{\partial u}{\partial x} = 2x - 2y
  \]

- \( v(x, y) = bx^2 - y^2 + 2xy \)
  \[
  \frac{\partial v}{\partial y} = 2xy - 2y + bx^2
  \]

Equating them from the first Cauchy-Riemann equation:
\[
2x - 2y = 2xy - 2y + bx^2
\]
Simplifying this:
\[
2x = 2xy + bx^2
\]
Dividing both sides by 2:
\[
x = xy + \frac{b}{2}x^2
\]
For this equation to hold true for all \(x\) and \(y\), we need \(b = 1\).

#### Second Cauchy-Riemann equation:
Now calculate the other partial derivatives.

- \( \frac{\partial u}{\partial y} = -2x + 2ay \)

- \( \frac{\partial v}{\partial x} = 2bx + 2y \)

From the second Cauchy-Riemann equation:
\[
-2x + 2ay = -2bx - 2y
\]
Substituting \(b = 1\):
\[
-2x + 2ay = -2x - 2y
\]
This simplifies to:
\[
2ay = -2y
\]
For this to hold for all \(y\), we must

Find a & b if f(z) = (x^2 - 2xy + ay^2) + i(bx^2 - y^2 + 2xy) is analytic. Hence find f(z) in terms of z.

Solve the problem of finding constants a and b such that f(z) = (x^2 - 2xy + ay^2) + i(bx^2 - y^2 + 2xy) is analytic by applying Cauchy-Riemann equations. Learn how to find f(z) in terms of z.

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