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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, $\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ & $\displaystyle \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$ 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?

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  • $\begingroup$ How can I show that $\lim \limits_{z \to 0} \frac{i\bar z^2}{z}$ doesn't exist ? $\endgroup$
    – Utshaw
    Aug 3, 2017 at 15:01
  • $\begingroup$ I was wrong, because the limit exists and is $0$. However, this doesn't mean the function is analytic, because the derivative only exists at a single point. $\endgroup$
    – egreg
    Aug 3, 2017 at 15:06
  • $\begingroup$ So , can I say that this function is analytic at a single point $(0,0)$ ? $\endgroup$
    – Utshaw
    Aug 3, 2017 at 15:08
  • $\begingroup$ It depends on how analytic is defined in your textbook, only you can check it. $\endgroup$
    – egreg
    Aug 3, 2017 at 15:09

4 Answers 4

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For $z=x+iy$ we have $f(z) = 2xy + i(x^2 - y^2) = i\bar{z}^2$ is not analytic.

Since $\partial_{\bar{z}}f \neq 0$.

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  • $\begingroup$ Please tell me what does $\partial_{\bar{z}}f \neq 0$ mean? $\endgroup$
    – Utshaw
    Aug 3, 2017 at 15:06
  • $\begingroup$ means it is not identicaly zero. since a complex function is analytic iff $\partial_{\bar{z}} f \equiv 0$. $\endgroup$
    – Guy Fsone
    Aug 3, 2017 at 16:49
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To be analytic in a point (e.g. in $(0,0)$) you need to verify CR on a neighborhood of that point. Here you have CR only at $(0,0)$ and not on a neighborhood of $(0,0)$. Hence not analytic.

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$f(z) = 2xy + i(x^2 - y^2)$

Cauchy conditions can be shortened as follows

$i{\partial f \over \partial x}={\partial f \over \partial y}$

${\partial f \over \partial x}=2 y+2 i x;\;{\partial f\over \partial y}=2 x-2 i y$

and as $\quad i(2 y+2 i x)\ne 2 x-2 i y\quad$ the function is not analytic

PS The function, written as a function of $z=x+iy$ is

$f(z)=\dfrac{i \left| z\right| ^4}{z^2}$

and appears less strange that it is not holomorphic

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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.

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  • $\begingroup$ I rechecked again & couldn't find any mistake . Is this function analytic only at $(0,0)$ ? $\endgroup$
    – Utshaw
    Aug 3, 2017 at 14:43
  • $\begingroup$ See @NAC answer... It's not analytic even at the origin. $\endgroup$ Aug 3, 2017 at 15:50

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