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It's hard to give any geometric intuition for the complex derivative since C is a 2-dimensional (real) vector space, so we have 4 real dimensions to try and interpret. My own interpretation is: at the very least, it's a measure of change - but I split the change into two components. The complex...

The derivative of a complex valued function is defined similarly to the real valued case: Let f be a real valued function. The the derivative of f at the point z0 is given by: f'(z) = \lim_{z \to z_0}\frac{f(z) - f(z_0)}{z - z_0} The only subtle difference is that z must be allowed to approach...

Fixed. There is also a nice proof on Wikipedia that only uses differentiation iirc. I'm wondering if the OP had the same concern when they were multiplying negative numbers? "You can't multiply a number a negative number of times ..."

The problem is that the square root function is multi-valued. sqrt(-1) = i or -i for instance. If you're always choosing i as "the" square root of negative 1, then 1 = sqrt(1) = sqrt(-1 * -1) != sqrt(-1) * sqrt(-1) = i * i = -1 is an obvious counterexample to the "rule" sqrt(ab) =...

It is constant 4 in the z-direction.

As t runs through [0,pi] the function makes a circle in the xy-plane at a hight of 4 in the z-direction

I think you're confusing topology and algebra. His proof is fine, though as Drongoski said the first two lines simply aren't necessary.

It's about the curves!

I would say \sqrt{\quad} : \mathbb{C} \to \{ z \in \mathbb{C} : \arg{z} \in [0, \pi) \} \cup \{0\} or \sqrt{\quad} : \mathbb{C} \to \{ z \in \mathbb{C} : \arg{z} \in [-\pi/2, \pi/2) \} \cup \{0\} or whatever, with the obvious definition (too lazy to type it out)

Domain and range is covered at high school, and I think you'll find definitions are more prevalent than you let on. Textbooks aren't written in a Definition / Theorem / Proof style, sure, but I'm sure most of the detail is there. Btw, there is no concept of "positive" when dealing with complex...

Implicit here is that x is in the domain of f. Nice. Your function h is not well-defined without choosing a branch of sqrt map. Otherwise, nice (assuming you apply the same observation about the domain/critical points as you did above). They definitely aren't useless, and definitions can't...

Cmon. Is -2 a critical point of sqrt(x)? Edit: This thread reminds me of the claim that 1/x is discontinuous at x=0 :(

Surely they say something about the domain of the original function? They don't consider 0 a critical point of 1/x do they?

Yes, after a comment like a definition is useless, only to follow by a (not so helpful) definition, you're missing an invitation to leave the mathematics forum for the mathematical crimes of (1) disparaging definitions, and (2) abuse of logic I'm happy to oblige. Joking .. mostly

A critical point of a function f is any point x in the domain of f for which the first derivative f' at the point x is 0 or not defined? I wouldn't call an inflection point a critical point (unless it's a horizontal inflection point), so don't take my word for it, take your text-book's/teacher's...

Show that \int_{0}^{2\pi} \frac{2}{3-2\sqrt{2}\cos\theta} \, d\theta = 4\pi

\begin{align*}(3+4i)^{-2} &= \left( \sqrt{3^2 + 4^2}e^{i\tan^{-1}(4/3)} \right)^{-2} \\ &= \frac{e^{-2i\tan^{-1}(4/3)}}{25} \\ \end{align*} Note: eix = cos(x) + isin(x) if you must.

Hahaha, I was like wtf is untouchablecuz asking this for?

\begin{align*} \sin(4t) + \sqrt{3}\cos(4t) = 0 &\Leftrightarrow \tan(4t) = -\sqrt{3} \\ &\Leftrightarrow 4t = \begin{cases} \frac{2\pi}{3} + 2k\pi, \ k \in \mathbb{Z} & \text{for } 4t \in \text{the second quadrant} \\ \frac{5\pi}{3} + 2k\pi, \ k \in \mathbb{Z} & \text{for } 4t \in \text{the...

A function sends one set of numbers to another set of numbers. We call these sets of numbers the domain and range respectively. For example, the function f defined by the relation f(x) = x2 sends some numbers (the x's) to some new numbers (the f(x)'s, or y's if you like). Which x's is this...