# Thread: Studying differentiability of modulus(z)^2

1. ## Studying differentiability of modulus(z)^2

Hello mathematicians,

I am asked to study the differentiability of lzl2 (complex analysis) and don't know how to do it :_(

I've read that it is only differentiable at 0, but how to reach that conclusion I ignore.

The two ways that I know for checking if a function is differentiable are

(1) By checking if the limit definition of derivative exists

But I don't know how to work with the squared modulus of numbers adding (numerator of the limit), and I don't think that's the way I'm supposed to do it...

(2) Checking where C-R equations hold. However, modulus(z)^2 doesn't have imaginary part, so I don't have u and v.

How can I study the differentiability of this function then?? I don't know to do this for f(z)=Re(z) either.

Thank you!

2. ## Re: Studying differentiability of modulus(z)^2

A function $\displaystyle f:\Bbb{C} \to \Bbb{R}$ is differentiable if and only if there exists a linear function $\displaystyle J:\Bbb{C} \to \Bbb{R}$ such that $\displaystyle \lim_{\Delta z \to 0} \dfrac{f(z+\Delta z) - f(z) - J(\Delta z)}{|\Delta z|} = 0$.

Edit: for the function $\displaystyle f:\Bbb{C} \to \Bbb{R}$ defined by $\displaystyle f(z) = |z|^2$, define $\displaystyle J_z(\Delta z) = J_z(\Delta x + i\Delta y) = 2x\Delta x + 2y\Delta y$ where $\displaystyle z = x+iy$. Then:\displaystyle \begin{align*}\lim_{\Delta z \to 0} \dfrac{|z+\Delta z|^2 - |z|^2 - J_z(\Delta z)}{|\Delta z|} & = \lim_{\Delta z \to 0} \dfrac{(x^2+2x\Delta x + (\Delta x)^2 + y^2+2y\Delta y + (\Delta y)^2) - (x^2+y^2) - (2x\Delta x + 2y\Delta y)}{|\Delta z|} \\ & = \lim_{\Delta z \to 0} \dfrac{|\Delta z|^2}{|\Delta z|} \\ & = \lim_{\Delta z \to 0} |\Delta z| = 0\end{align*}

3. ## Re: Studying differentiability of modulus(z)^2

Originally Posted by Phyba
Hello mathematicians,

I am asked to study the differentiability of lzl2 (complex analysis) and don't know how to do it :_(

<snip>

(2) Checking where C-R equations hold. However, modulus(z)^2 doesn't have imaginary part, so I don't have u and v.

How can I study the differentiability of this function then?? I don't know to do this for f(z)=Re(z) either.

Thank you!
$w=u+\imath v=|z| = |x+\imath y| = \sqrt{x^2(t)+y^2(t)}$

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

$v=0$

$u_x=\dfrac{x}{\sqrt{x^2+y^2}}$

$u_y=\dfrac{y}{\sqrt{x^2+y^2}}$

$v_x = v_y = 0$

clearly the CR equations are not satisfied anywhere but at $z=0$

4. ## Re: Studying differentiability of modulus(z)^2

For $\displaystyle f(z) = \text{Re}(z)$, you can do the same thing. Breaking it down component-wise, you have $\displaystyle f(x+iy) = x$, so $\displaystyle \dfrac{\partial f}{\partial x} = 1$ and $\displaystyle \dfrac{\partial f}{\partial y} = 0$. Then:

$\displaystyle \lim_{\Delta z \to 0} \dfrac{f(z+\Delta z) - f(z) - \begin{bmatrix}1 & 0\end{bmatrix}\begin{bmatrix}\Delta x \\ \Delta y\end{bmatrix} }{|\Delta z|} = \lim_{\Delta z \to 0} \dfrac{x+\Delta x - x - \Delta x}{|\Delta z|} = 0$

5. ## Re: Studying differentiability of modulus(z)^2

Thank you very much!

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# show that f (z) =lzl 2 is differentiable only at z=0

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