I am preparing to take my first course in complex variables. I am reading some lecture notes online. They claim that the function $f(z) = Re(z)$ is continuous but NOT differentiable. I know the definitions of a limit, of continuity, and of a derivative. I understand why this function is continuous. I am trying to show that it is NOT differentiable.
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- 1$\begingroup$ Try to calculate $f'(1)$. $\endgroup$user228113– user2281132016-05-17 21:55:51 +00:00Commented May 17, 2016 at 21:55
- 3$\begingroup$ Better to say Re z is not complex differentiable. $\endgroup$zhw.– zhw.2016-05-17 23:52:36 +00:00Commented May 17, 2016 at 23:52
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4 Answers
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1 The definition of derivative can be written as
$$ f'(z) = \lim_{h \to 0} \dfrac{f(z+h) - f(z)}{h} $$
which looks just like the real-variable definition, but here this is taken in the complex sense, i.e. $h$ is allowed to be a complex number. $h \to 0$ means the distance in the complex plane from $h$ to $0$ goes to $0$ (or equivalently, both real and imaginary parts of $h$ go to $0$). In order for the limit to exist, you must always get the same value as $h \to 0$ in any manner.
If you take $h$ to be real, $f(z+h) = f(z) + h$ and the quotient is $1$.
If you take $h$ to be imaginary, $f(z+h) = f(z)$ and the quotient is $0$.
The limit as $h \to 0$ doesn't exist: it can't be both $1$ and $0$. Thus we say $f'(z)$ doesn't exist, and the function is not differentiable.
- 1$\begingroup$ More informative and intuitevely plausible answer for a novice than mine. (+1)! $\endgroup$MathematicianByMistake– MathematicianByMistake2016-05-17 22:00:46 +00:00Commented May 17, 2016 at 22:00
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For a function to be differentiable in $\Bbb{C}$, it must satisfy the Cauchy-Riemann equations, that is, if $$f(x,y)=u(x,y)+iv(x,y)$$ it must satisfy $$u_x=v_y\\u_y=-v_x$$
But for $f(z)=\Re(z)=x$ we get $$u_x=1\neq v_y=0$$
So it is not differentiable.
answered May 17, 2016 at 21:51
MathematicianByMistake
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Another way to see it, it is that the real part of a complex number can be written with its conjugate: $Re(x) = \frac{1}{2} (x + x^*)$. Since the conjugate function is the classical example of a non-complex-differentiable function (see for exampe this), it follows that the real part is not complex-differentiable.
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@Zaccharie Ramzi. the link is not working.
What i thought of is:
Sine: Re(z) = (z + z*)/2.
Define: g,h,k to be: z -> z, z -> z*, z -> 1/2.
Then: Re = g+h/k, but k'= 0 for all z. so it is not diff'able
