Check if $f(z) = 2xy + i(x^2 - y^2) $ is analytic or not Check whether the function defined by $f(z) = 2xy + i(x^2 - y^2) $ is analytic or not.

Sufficient condition for a function being analytic: The sufficient condition for a function $f(z) = u + iv$ to be analytic at all the points in region $R$ are
$\displaystyle (a)\; \frac{\partial u}{\partial x}  = \frac{\partial v}{\partial y} ,\frac{\partial u}{\partial y}  = -\frac{\partial v}{\partial x} $ 
$\displaystyle (b)\; \frac{\partial u}{\partial x}  , \frac{\partial v}{\partial y} ,\frac{\partial u}{\partial y}  ,\frac{\partial v}{\partial x} $ are continuous functions of $x$ and $y$ in region $R$ .
Here, $u=2xy$ , $v = x^2-y^2$
$\displaystyle \frac{\partial u}{\partial x} = 2y$,
$\displaystyle \frac{\partial v}{\partial y} = -2y$
$\displaystyle \frac{\partial u}{\partial y} = 2x$,
$\displaystyle \frac{\partial v}{\partial x} = 2x$
If this is analytic function then,
$\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ &amp; $\displaystyle \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$
So,$x = y = 0$The partial derivatives are continuous .
So, I think function is analytic only at $(0,0)$ but the answer says that it is not analytic.Am I wrong or the answer in the book is incorrect?
For $z=x+iy$  we have $f(z) = 2xy + i(x^2 - y^2) = i\bar{z}^2$  is not analytic.

Since $\partial_{\bar{z}}f \neq 0$.
Hint: You've concluded that for the C-R equations to hold, you need to have $x = y= 0$. So at $x = 1, y = 0$, is the function analytic? No. 

Alternatively, you may have made a mistake in checking the C-R equations. I leave it to you to decide which.