$$u = x^2 + 2axy + by^2 \hspace{1cm} v = dx^2 + 2cxy + y^2$$ $$\frac{\partial u}{\partial x} = 2x + 2ay \hspace{1cm} \frac{\partial v}{\partial x} = 2dx + 2cy$$ $$\frac{\partial u}{\partial x} = \hspace{1cm} \frac{\partial v}{\partial y} = 2cx + 2y$$

But f(z) is analytic

$\therefore$ ux = vy & uy = -vx

2x + 2ay = 2cx + 2y (1) ( Comparing the coefficients of x & y)

a = 1, c = 1

Similarly, uy = -vx

2ax + 2by = -(2dx + 2cy)

$\therefore$ 2x + 2by = -2dx - 2y

Comparing the coefficients of x& y, v + iu, we get

b = -1 & d = -1

Differentiating v w.r.t. x & y we get

$$v_{a} = \frac{(x^2 + y^2) \frac{d}{dx}x + x\frac{d}{dx}(x^2 + y^2)}{(x^2 + y^2)^2} + cosh y(-sinx)$$ $$v_{x} = \frac{-x^2 + y^2}{(x^2 + y^2)^2} - sinxcoshy$$ $$v_{y} = \frac{-2xy}{(x^2 + y^2)^2} + cosxsinhy$$

According to the definitin of analytic

$u_{x} = v_{y}$ & $u_{y} = -v_{x}$

Now by Milne Thompson,

$\phi_{1} = v_{y}[z,0] = 0$ i.e. x = z & y = 0

$\phi_{2} = v_{x}[z,0] = \frac{-z^2}{z^4}-sinz$

$f'(z) = \phi_{1} + i\phi_{2} = i\bigg[\frac{-z^2}{z^4} - sinz\bigg]$

Integrating w.r.t. z we get

$f(z) = i \int \bigg[\frac{-1}{z}-sinz \bigg]dz = i \bigg[\frac{-1}{-z} + cosz \bigg] + c = i \bigg[\frac{1}{z} + cosz \bigg]+c$