I have questions on Part II of the same exercise asked here: $f(z)=u(x,y)+iv(x,y), u$ and $v$ are differentiable. Is $f$ differentiable?
$\label{1}{1}$. Is this ambiguous, but probably meant to say something like
if $u$ and $v$ are $\mathbb R^2-$ differentiable at a point $\mathbb R^2 \ni (x,y)$, then f is $\mathbb C-$ differentiable at the corresponding point point $\mathbb C \ni z=x+iy $
?
I hadn't considered $\mathbb R-$ vs $\mathbb C-$ differentiable.
$\label{2}{2}$. Is $f$ $\mathbb C-$differentiable if $u$ and $v$ are if $u$ and $v$ are $\mathbb C-$differentiable?
I think yes. Here is my reasoning. What are the errors, if any?
$$f'(z) := \lim_{h \to 0}\frac{f(z+h)-f(z)}{h} = \lim_{h \to 0}\frac{u(z+h)+iv(z+h)-u(z)-iv(z)}{h}$$
$$ = \lim_{h \to 0}\frac{u(z+h)-u(z)}{h} + i\lim_{h \to 0}\frac{v(z+h)-v(z)}{h} = u'(z)+iv'(z)$$
$\therefore, f$ is $\mathbb C$-differentiable at $z_0$ if $u$ and $v$ are complex differentiable. Furthermore, the $\mathbb C$-differentiability of $f$ is concluded by Definition (see end) and not by Cauchy-Riemann Thm 2.13(b) (see end). QED
$\label{3}{3}$. If $u$ and $v$ are complex differentiable, then do we have $f'(z)=u'(z)=v'(z)=0$ ?
I think yes. Here is my reasoning. What are the errors, if any?
I also hadn't considered that $u$ and $v$ are actually $\mathbb R$-valued. I cannot apply Exer 2.19 (see below), but by similar reasoning, I deduce that with Cauchy-Riemann Thm 2.13(a) and by the complex differentiability of $u$ at $z$, $u'(z)=0$. Similarly, $v'(z)=0$. $\therefore, f'(z)=0+i0=0.$ QED
$\label{4}{4}$. Is $f$ $\mathbb R^2-$differentiable if $u$ and $v$ are $\mathbb R^2-$differentiable?
It seems yes based on other question, but how would you prove it? Here is my reasoning. What are the errors, if any?
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4.1. This book doesn't seem to have an explicit definition of $\mathbb R^2-$differentiability of a $\mathbb C$-function based on my other question, but I guess it's to do with treating $f=f(x,y)$ and $i$ as if $i$ were a $\mathbb R$-constant.
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4.2. I thought to go back to the $\mathbb R^2-$differentiability in Stewart Calculus 14.4.7, but if $i$ is indeed treated as a $\mathbb R$-constant, then $f$ is a linear combination of $\mathbb R^2-$differentiable functions and $\therefore$ $\mathbb R^2-$differentiable.
Note: 4.1 is about definition. 4.2. is about the argument assuming my definition is the same as the book's reasonably deduced implicit definition, the book's explicit definition that I overlooked or some widely accepted definition of $R^2-$differentiability of a $\mathbb C$-function.
$\label{5}{5}$. Actually, $f$ is $\mathbb R^2-$differentiable only if $u$ and $v$ are $\mathbb R^2$-differentiable, by similar reasoning as in 4 whence
f is $\mathbb R^2$-differentiable $\iff$ $\Re(f)$ and $\Im(f)$ are $\mathbb R^2$-differentiable
?
$\label{6}{6}$. Is $f$ not necessarily $\mathbb C-$differentiable if $u$ and $v$ are if $u$ and $v$ are $\mathbb R^2-$differentiable?
Okay, so the answer is yes, answered in the aforementioned question again linked here. Here is my reasoning. What are the errors, if any?
As in the aforementioned question, $f(z)=\overline{z}=x-iy$ is not $\mathbb C-$differentiable. Instead of proving this directly with Cauchy-Riemann Thm 2.13(b), I will prove by contradiction and by considering $f$'s components:
Argument 1: If we try to write its derivative as $f'(z)=u'(z)+iv'(z)$, we see $f'(z)$ dne because the two addends $u'(z)$ and $iv'(z)$ do not exist because $u(z)$ and $v(z)$, while $\mathbb R^2$-differentiable, are not $\mathbb C$ differentiable, this time deduced by Cauchy-Riemann Thm 2.13(b). QED
Argument 1 rephrased: If we try to write $\overline{z}$'s derivative as $(\overline{z})'=(\Re(\overline{z}))'+i(\Im(\overline{z}))'$, we see $(\overline{z})'$ dne because the two addends $(\Re(\overline{z}))'$ and $i(\Im(\overline{z}))'$ do not exist because $\Re(\overline{z})$ and $\Im(\overline{z})$, while $\mathbb R^2$-differentiable, are not $\mathbb C$ differentiable, this time deduced by Cauchy-Riemann Thm 2.13(b). QED
Argument 2: (Same as Argument 1 except the last phrase) If we try to write $\overline{z}$'s derivative as $(\overline{z})'=(\Re(\overline{z}))'+i(\Im(\overline{z}))'$, we see $(\overline{z})'$ dne because the two addends $(\Re(\overline{z}))'$ and $i(\Im(\overline{z}))'$ do not exist because $\Re(\overline{z})=\Re(z)$ and $\Im(\overline{z})=-\Im(z)$, while $\mathbb R^2$-differentiable, are not $\mathbb C$ differentiable, because they don't satisfy C-R…which is a vacuous truth because they do satisfy C-R. Instead, we can prove by definition: $$\lim_{h \to 0} \frac{\Re(z+h)-\Re(z)}{h} = \lim_{h \to 0} \frac{h_x}{h_x+ihy}$$
This does not exist because along $h_x=0$, $\lim=0$ while along $h_y=0$, $\lim=1$.
Similarly,
$$\lim_{h \to 0} \frac{-\Im(z+h)+\Im(z)}{h} = \lim_{h \to 0} \frac{-h_y}{h_x+ihy}$$
This does not exist because along $h_y=0$, $\lim=0$ while along $h_x=0$, $\lim=i$.
Argument 3: Same as argument 2 except that I say that the parts are not complex differentiable and hence $f$ is not complex differentiable, but I see why is this wrong as well.
$\label{7}{7}$. If Reasoning in 6 (Argument 1) is right, then we actually have additionally proved (hope none of those modifiers were misplaced!) that
f is $\mathbb C$-differentiable $\iff$ $\Re(f)$ and $\Im(f)$ are $\mathbb C$-differentiable
?
(Note: Vacuous truths acceptable! Hehe) –> i.e. Update: Reasoning in 6 was (Argument 1 is) wrong, conclusion is wrong, but conclusion is right if Argument 1 is right! Hehe
Definition of $\mathbb C$-differentiability
Cauchy-Riemann Thm 2.13(a)(b)
Exer 2.19
Best Answer
$(1)$ Probably not: $u$ and $v$ are complex differentiable iff they are constant. $f$ is obviously differentiable in $\mathbb R^2$ or $\mathbb C$ if its parts are, RESP, differentiable in $\mathbb R^2$ or $\mathbb C$
(2), (3) Yes albeit trivially.
(4)-(7)