If a complex-valued function is continuous, are the component real and imaginary parts $u(x,y)$ and $u(x,y)$ necessarily continuous? If so, why?
[Math] Continuity of the real and imaginary parts of a continuus complex-valued function
complex-analysis
Related Solutions
There is another approach to this. First we can define limits for a complex valued function of a real variable:
Let $c$ be a real number and let $f$ be a complex valued function of a real variable defined in a certain deleted neighborhood of $c$. In other words there exists an open interval $I$ such that $c\in I$ and $f:I\setminus{c} \to\mathbb{C} $ be a function. A complex number $L$ is said to be the limit of $f$ at $c$ (denoted by $\lim_{x\to c} f(x) =L$ or $f(x) \to L$ as $x\to c$) if for any real number $\epsilon >0$ there exists a corresponding real number $\delta>0$ such that $|f(x) - L|<\epsilon $ whenever $0<|x-c|<\delta$.
Next comes the derivative:
Let $I$ be an open interval and $c\in I$ and $f:I\to\mathbb{C}$ be a function. The derivative of $f$ at $c$ denoted by $f'(c) $ is defined by the equation $$f'(c) =\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}$$ and $f$ is said to be differentiable at $c$ if the above limit exists (in the above equation $h$ is a real number).
These definitions are exactly the same as those for real valued functions of a real variable. It can be easily proved (in an almost obvious fashion) that the above definition of derivative of a complex valued function of a real variable is equivalent to the definition of derivative given in your question. Depending on the way $f$ is defined one of the definitions may be preferable / convenient to use. Thus if real and imaginary parts of $f$ are easily evaluated then it makes sense to use the definition in your question. If the real and imaginary parts of $f$ are not available easily then one can use the definition in this answer. Note that all the usual rules of differentiation are valid in this scenario and can be established using both the definitions with equal ease.
The example $f(x)=e^{ix}$ in your question is very interesting. Let's define it via power series for $e^{z}$ as $$e^{z} = 1+z+\frac {z^{2}}{2!}+\frac{z^{3}}{3!}+\dots,z\in\mathbb {C} \tag{1}$$ Using binomial theorem for positive integer index and Cauchy's product formula for multiplication of two infinite series it can easily proved that $$e^{z+w}=e^{z}e^{w},z,w\in\mathbb {C}\tag{2}$$ Using this definition of $e^{z} $ we observe that the real and imaginary parts of $f(x) =e^{ix} $ are not available in a simple form (they are just available as infinite series) and hence we can apply the definition given above to evaluate the derivative $f'(x) $ as follows \begin{align} f'(x) & =\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\notag\\ &=\lim_{h\to 0}\frac{e^{i(x+h)}-e^{ix}}{h}\notag\\ &=\lim_{h\to 0}e^{ix}\cdot\frac{e^{ih}-1}{h}\notag\\ &=e^{ix}\lim_{h\to 0}\frac{e^{ih}-1}{h}\tag{3} \end{align} Next note that \begin{align} \left|\frac{e^{ih} - 1}{h}-i\right|&=\left|\frac{i^{2}h}{2!}+\frac{i^{3}h^{2}}{3!}+\dots\right|\notag\\ &\leq \frac{|h|} {2!}+\frac{|h|^{2}}{3!}+\dots\notag\\ &<\frac{|h|} {2}+\frac{|h|^{2}}{2^{2}}+\dots \notag\\ &=\frac{|h|} {2-|h|}\text{ (provided} |h|<2)\notag \end{align} By Squeeze Theorem we now get $$\lim_{h\to 0}\left|\frac{e^{ih}-1}{h}-i\right|=0$$ and hence $$\lim_{h\to 0}\frac{e^{ih}-1}{h}=i$$ Now from equation $(3)$ we get $$f'(x) =if(x) \tag{4}$$ Thus the usual rule for derivative of $f$ is a consequence of the definition of $f$ and the definition of derivative. It is not automatic as some may believe.
On the other hand the function $g(x) =\cos x-i\sin x$ is defined in terms of its real and imaginary parts and hence applying the definition in your question we get $$g'(x) =- \sin x-i\cos x=-i(\cos x-i\sin x) =-ig(x) \tag{5}$$ Consider the function $h(x) =f(x) g(x) $ whose derivative is $$h'(x) =f'(x) g(x) +f(x) g'(x) =if(x) g(x) - if(x) g(x) =0$$ Therefore $h$ is constant and $$h(x) =h(0)=f(0)g(0)=1\cdot 1=1$$ It follows that $$e^{ix} =f(x) =\frac{1}{g(x)}=\frac{1}{\cos x-i\sin x} =\cos x+i\sin x\tag{6}$$ for all real values of $x$. Equating real and imaginary parts we get the familiar series for circular functions and this constitutes a proof of these series without using Taylor's theorem (the proof uses multiplication property of exponential series given in $(2)$ to get derivative of $f$ and derivatives of circular functions to get derivative of $g$).
Note: The mean value theorem does not apply for complex valued functions of a real variable, but the result that "vanishing of derivative implies the function is constant" applies here too (this is proved by applying the result to real and imaginary parts separately) and has been used in the above proof.
It is common to misunderstand "$u$ is a function of $x$ and $y$" as "$u$ depends on $x$ and $y$". Which leads to errors such as thinking that $u(x,y)=x^2$ is not a function of $x $ and $y$.
Saying that "$u$ is a function of $x$ and $y$" means that the domain of $u$ is the set of ordered pairs $(x,y)$. Informally, $u$ takes two real numbers as inputs, and produces a number as output. To qualify as a function, $u$ must be consistent: if given the same pair $(x,y)$ today and tomorrow, it must give the same output. Other than that, it's free to use the inputs in any way it wants; and that includes not using them at all. For example, $f(x,y)=\sqrt{2}$ is a function which always returns $\sqrt{2}$ as output. It's still a function of $x$ and $y$ -- meaning it takes $x$ and $y$ as inputs -- it just does not use those inputs.
Best Answer
The functions $\operatorname{Re}, \operatorname{Im}: \mathbb{C} \to \mathbb{R}$ are continuous since $\operatorname{Re} (z_1+z_2) = \operatorname{Re} z_1 + \operatorname{Re} z_2$ and $|\operatorname{Re} z | \le |z|$, and similarly for $\operatorname{Im}$.
Hence $\operatorname{Re} \circ f$ and $\operatorname{Im} \circ f$ are continuous.