1

Let $f: \mathbb{R} \to \mathbb{R}$ be a real function and let $z \in \mathbb{C}$ be a complex number such that $$ f(x)=|x \cdot z| $$ Let's calculate the derivative of $f$

if we applicate the derivation rules: $$ f'(x)=\dfrac{x \cdot z}{|x \cdot z|} \cdot z $$ but it's wrong indeed $$ f(x)=|x \cdot z| = |x| \cdot |z| $$ and now $$ f'(x)=\dfrac{x}{|x|} \cdot |z| $$ so what's the derivative of $f$?

In general what's the derivative of absolute value of a function $|f(x,z)|$ respect the real variable $x$ and $z \in \mathbb{C}$?

Thanks.

Matey Math
  • 1,701
  • I'm not sure you actually mean to say "Let $f : \mathbb{R} \to \mathbb{R}$ be a real function" and afterwards pick a complex $z$ such that $f(x) = | x \cdot z|$, for this is obviously not possible for all possible $f$. – SvanN Jul 09 '17 at 11:46
  • Apparently what you mean to say is "let z be a complex number and then define f:R→R by $f(x)= |xz|$. – user247327 Jul 09 '17 at 13:06
  • @user247327 i mean $f$ is a real function because even though $z$ is complex but $|xz|$ is real – Matey Math Jul 09 '17 at 13:10

3 Answers3

8

I'm going to deal with the general problem: Given a complex valued function $$g:\quad{\mathbb R}\to {\mathbb C},\qquad x\mapsto g(x)=u(x)+i v(x)\ ,$$ one has $|g(x)|=\sqrt{u^2(x)+v^2(x)}$ and $g'=u'+i v'$. Therefore $${d\over dx}\bigl|g(x)\bigr|={u(x)u'(x)+v(x)v'(x)\over\sqrt{u^2(x)+v^2(x)}}={{\rm Re}\bigl(g(x) \overline{ g'(x)}\bigr)\over|g(x)|}\ .\tag{1}$$ In the example at hand $z$ is a constant, and $g(x):=xz$, so that $g'(x)=z$. According to $(1)$ one then has $${d\over dx}\bigl|x\,z\bigr|={{\rm Re}\bigl(xz\,\bar z\bigr)\over|x\,z|}={x\,|z|^2\over |x|\,|z|}={x\over|x|}\,|z|\qquad\bigl(xz\ne0)\ .$$

Christian Blatter
  • 226,825
  • ok @Christian Blatter thanks so much, that's just what i needed – Matey Math Jul 09 '17 at 13:57
  • The symbol dash above g'(x) function in equation (1) usually refers to negation. But the answer meant complex conjugate or negation of imaginary part only so the symbol should be *. – Aschoolar Apr 03 '21 at 15:59
  • @Aschoolar: The "\overline" (e.g., $\overline{g'(x)}$) is standard notation for the complex conjugate. In analysis we seldom have to express "negations" applied to expressions. – Christian Blatter Apr 03 '21 at 18:31
1

Interesting. We try to use the chain rule I guess.

You are using the formula $$ \frac{d}{dx} |x| = \frac{x}{|x|} \tag{*}$$ which is true when $x$ is real and nonzero. Then you try to use the chain rule like this $$ \frac{d}{dx} \big|g(x)\big| = \frac{g(x)}{|g(x)|}\;g'(x) \tag{**}$$ This will be OK if $g$ is a differentiable function whose values are nonzero reals. But you attempt to apply it to the function $g(x) = xz$ with non-real values. No good. As I said, (*) is only for real values, so (**) is only for real-valued functions $g$.

GEdgar
  • 111,679
  • you say $g(x)=xz$ but if i define $f(x)=|g(x)|$ then $f$ is a real function so the problem remains – Matey Math Jul 09 '17 at 13:09
  • $g(x)$ is not a real function, $g(x)$ is the function you have inside the chain rule using $()$, and that is what you cannot* have in $(**)$. – GEdgar Jul 09 '17 at 13:11
  • @GEdagar $g(x)$ is not a real function it's sure but $f(x)=|g(x)|$ it's a real function and i want to calculate the derivative of $f$ – Matey Math Jul 09 '17 at 13:14
1

$f(x)=|x\cdot z|$ is a real function, for any $z\in\mathbb{C}$

Let $z= a+bi;\;a,b\in\mathbb{R}$

then $$f(x)=|x\,z|=|ax+bxi|=\sqrt{a^2x^2+b^2x^2}$$ and $$f'(x)=\frac{a^2 x+ b^2 x}{ \sqrt{a^2 x^2+b^2 x^2}}=|z|\cdot \dfrac{x}{|x|}$$

Raffaele
  • 26,371
  • it's clear that $f'(x)$ is that, but my question is why the derivative rules doesn't work well? how derivate $|f(x,z)|$ in general? – Matey Math Jul 09 '17 at 13:18