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I am trying to check if the function $f(z)=z+|z|$ is analytic by using the Cauchy-Riemann equation.

I made $z = x +jy$

and therefore

$$f(z)= (x + jy) + \sqrt{x^2 + y^2}$$

put into $f(z) = u+ jv$ form:

$$f(z)= x + \sqrt{x^2 + y^2} + jy$$

where

$u = x + \sqrt{x^2 + y^2}$

and that

$v = y$

Now I need to apply the Cauchy-Riemann equation, but don't know how would I go about doing that.

Any help would be much appreciated.

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You just need to check whether or not the CR equations are true in this case. That means compute $\partial u/\partial x, \partial u/\partial y, \partial v/\partial x,\partial v/\partial y$, and then see whether or not the equalities CR claims holds between these quantities actually holds. –  anon Jun 19 '12 at 6:14
1  
Since the difference of two analytic functions is analytic, it is enough to check $\mid z \mid$ for analyticity. –  Georges Elencwajg Jun 19 '12 at 6:17
    
The function $g$ with $g(z)=|z|$ alone is not differentiable. For any nonzero $z_0$, motion in the direction concentric to the origin will leave $g$ constant, while motion in the radial direction will change $g(z)$ like a linear function. Thus $g'(z)$ would want to be $0$ and some nonzero real number at the same time. –  alex.jordan Jun 19 '12 at 6:25

2 Answers 2

up vote 1 down vote accepted

The Cauchy-Riemann equations are \begin{align} \dfrac{\partial u}{\partial x} & = \dfrac{\partial v}{\partial y}\\ \dfrac{\partial v}{\partial x} &= -\dfrac{\partial u}{\partial y} \end{align} In your case, $u(x,y) = x + \sqrt{x^2+y^2}$ and $v(x,y) = y$. Assuming $(x,y) \neq (0,0)$, the partial derivatives are \begin{align} \dfrac{\partial u}{\partial x} & = 1 + \dfrac{x}{\sqrt{x^2+y^2}}\\ \dfrac{\partial v}{\partial x} & = 0\\ \dfrac{\partial u}{\partial y} & = \dfrac{y}{\sqrt{x^2+y^2}}\\ \dfrac{\partial v}{\partial y} & = 1 \end{align} Hence, from the Cauchy-Riemann equations, we get that $$1 + \dfrac{x}{\sqrt{x^2+y^2}} = 1 \implies \dfrac{x}{\sqrt{x^2+y^2}} = 0$$ $$\dfrac{y}{\sqrt{x^2+y^2}} = 0$$ This has no solutions since $(x,y) \neq (0,0)$. Hence, the function is not differentiable on $\mathbb{C} \backslash \{(0,0)\}$. The only point we need to check whether it is differentiable is $(0,0)$. At this point, we can check for differentiability directly from the definition. You will find that it is also not differentiable at $(0,0)$. Hence, the function is nowhere analytic.

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In order for your function to be analytic, it must satisfy the Cauchy-Riemann equations (right? it's good to think about why this is true). So, what are the equations?

Well, du/dx = dv/dy.

Does this hold? Or you could consider du/dy = -dv/dx.

If either of these equations do not hold, then the function is not analytic.

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Note that by not analytic, I mean not analytic everywhere, which I think is what you mean as well. –  MGN Jun 19 '12 at 6:22

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