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Let $f(x+iy) = x^2 + i.0$.

Then $u(x,y) = x^2$ and $v(x,y) = 0.$

Hence $u_x = 2x$, $v_y = 0$, $u_y = 0$, and $v_x = 0.$

Clearly this doesn't satisfy Cauchy Riemann equations, and hence is not differentiable.

But we know from calculus that $f'(x) = 2x.$ So my question is, where I went wrong in my analysis?

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    You have $f(z) = (Rez)^2$ but Rez is not differentiable. Another thing is what $f'(z)$ is even if you know $f'(x) = 2x$2012-10-21

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There is a difference between real differentiation and complex differentiation. The function $f(x+iy)=x^2$ is real differentiable as a function of $x$ (with derivative $2x$) but, as you showed from applying the Cauchy-Riemann equations, is not complex differentiable as a function of $z=x+iy$.

The question is: why is a function that is real differentiable not necessarily complex differentiable (as you've shown)? Recall that the derivative of a function $f(x)$ is defined as $ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}. $ If $f$ is function of a real number then $h$ has to be real. Roughly speaking, the real number $h$ can only approach the limit 0 from two directions. Either you let $h$ be positive and make it smaller and smaller, or you let $h$ be negative and make it larger and larger, so approaching 0 in two ways. For the limit to exist these two ways of approaching 0 have to give the same result. This is why the function $f(x)=|x|$ is not differentiable at $x=0$: the result depends on whether you start with $h$ negative (giving "$f'(x)=-1$") or $h$ positive (giving "$f'(x)=+1$").

On the other hand, if $f$ is a function of a complex variable then $h$ is also complex. If $h$ is complex then it can approach 0 in an infinite number of ways. Think of the two dimensional complex plane: you can travel towards the origin $z=0$ though any of the infinite number of lines that pass through $z=0$. In particular, we could let $h$ approach 0 along the imaginary axis by setting $h=it$ for a real number $t$.

Let's try this out directly and show that your function $f(x+iy)=x^2+0i$ is not complex differentiable at $z=x+iy=1$. If we let $h$ be real then we get $ f'(1) = \lim_{h\rightarrow 0}\frac{f(1+h)-f(1)}{h}\\ f'(1) =\lim_{h\rightarrow 0}\frac{(1+h)^2-(1)^2}{h}\\ f'(1) =\lim_{h\rightarrow 0}\frac{1+2h+h^2-(1)^2}{h}\\ f'(1) =\lim_{h\rightarrow 0}(2+h) = 2 $ On the other hand, if we let $h$ be purely imaginary by setting $h=it$ for a real $t$ we get $ f'(1) = \lim_{h\rightarrow 0}\frac{f(1+it)-f(1)}{h}\\ f'(1) =\lim_{h\rightarrow 0}\frac{(1)^2+0t-(1)^2}{h}\\ f'(1) =\lim_{h\rightarrow 0}\frac{0}{h}=0 $ Thus when we allow $h$ to be complex it is possible for the limit defining the derivative to take on different values depending on the way in which you approach 0, and hence the limit doesn't exist.

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    Great answer! I think that this is such a great way to help develop more of an intuition about the Cauchy-Riemann equations and the key differences between real and complex analysis. Wish this had been taught in my class to lay the foundation for the more rigorous proof we went through!2018-05-07
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The function you're describing is $f(z)=Re(z)^{2}$. Even if you break a function into its real and imaginary parts, you have to bear in mind that you're still dealing with a complex function - you can't just ignore that and say $f(x)=2x$, as that's only true for a subset of the domain of $f$.

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It looks like you're trying to think of $f(x+iy)=x^2$ in a way similar to $f(z)=z^2$, where $f'(z)=2z$. Your calculations are correct though.

Another way to approach complex differentiability is to see if the function depends on the conjugate $\bar{z}$. One nice result is that a function is (complex) differentiable if and only if $\frac{\partial f}{\partial \bar{z} }=0$. Since $x=\frac{z+ \bar{z}}{2}$ your function can be expressed as $f(x+iy)= \left( \frac{z+ \bar{z}}{2} \right)^2$, which depends on $\bar{z}$.