0
$\begingroup$

Let $\gamma(t):[a,b]\rightarrow \mathbb{C}$ , $[a,b]\subset \mathbb{R}$, $\gamma(t)= \gamma_1(t) + i\gamma_2(t)$

$\int_{a}^{b} \gamma(t) dt = \int_{a}^{b} \gamma_1(t) dt + i\int_{a}^{b} \gamma_2(t) dt$ and $\frac{d}{dt} \gamma(t)= \frac{d}{dt} \gamma_1(t)+ i\frac{d}{dt} \gamma_2(t)$

When differentiating and integrating $\gamma_1(t)$ or $\gamma_2(t)$ with respect to $t$, should i treat $\gamma_1$ / $\gamma_2$ as a real-valued function and any complex constant in the formula as a real constant? Then the integration and differentiation rules learned in Calculus will apply nicely.

Please help me by giving explanation on why or why not. Thanks!

For example, $\int_{0}^{\pi}e^{i2t}dt = \bigg[\dfrac{e^{i2t}}{2i}\bigg]^{0}_{\pi}$ . Here, i treat $2i$ as a constant and do the normal integration rule as i have learnt in Calculus.

$\frac{d}{dt}\big[ \sin(3it)\big] =3i\cdot \cos(3it)$. Here, i treat $3i$ as constant and do the normal chain rule as i have learnt in Calculus.

$\endgroup$

closed as unclear what you're asking by Siong Thye Goh, user91500, Ennar, Namaste, José Carlos Santos Sep 23 '17 at 19:49

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ That's at least how integrating works. I'm pretty sure it's also how differentiating works, but I'm a little rusty. $\endgroup$ – themathandlanguagetutor Sep 23 '17 at 0:39
  • $\begingroup$ Can you explain further on why it works? thanks $\endgroup$ – Little Rookie Sep 23 '17 at 1:39
2
$\begingroup$

The simple answer is that the 'usual facts' about derivatives for real functions do not automatically transfer to complex functions. For example, $\ln(|z|)$ does not have a complex derivative with respect to $z$. So your question is a good one; since you are not assuming that properties for reals carry over to complex numbers. (See this post for another common kind of error due to such assumptions.) $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

So what holds? If you build a more general framework for differentiation, such as sketched in this post, you can establish many basic properties of derivatives, of which I shall list some below. This framework is not only more intuitive, it also clearly shows why complex differentiability is much stronger than real differentiability.

Specifically, given any real variables $x,y$ varying with real parameter $t$, for $\lfrac{dx}{dy}$ to be defined when $t = p$ we merely need $\lfrac{Δx}{Δy}$ to have a limit as $t \to p$. This is along a path. In contrast, given complex variables $x,y$ varying with complex parameter $t$, for $\lfrac{dx}{dy}$ to be defined when $t = p$ we need $\lfrac{Δx}{Δy}$ to have a limit as $t \to p$ along all possible paths.

But since the definition itself does not care what type the parameter is, nor what type the variables are, except that the values come from a field with a norm, all the basic properties that arise from algebraic identities hold with the same proof. Time for the examples.


Given any complex variables $x,y,z$ varying with some parameter $t$:

  • $\lfrac{dc}{dx} = 0$ whenever $\lfrac{dc}{dx}$ is defined, for any constant $c$.

  • $\lfrac{d(c·y)}{dx} = c·\lfrac{dy}{dx}$ whenever $\lfrac{dy}{dx}$ is defined, for any constant $c$.

  • $\lfrac{d(y+z)}{dx} = \lfrac{dy}{dx}+\lfrac{dz}{dx}$ whenever $\lfrac{dy}{dx},\lfrac{dz}{dx}$ are both defined.

  • $\lfrac{dx}{dz} = \lfrac{dx}{dy}·\lfrac{dy}{dz}$ whenever $\lfrac{dx}{dy},\lfrac{dy}{dz}$ are both defined.

  • $\lfrac{d(y·z)}{dx} = \lfrac{dy}{dx}·z+y·\lfrac{dz}{dx}$ whenever $\lfrac{dy}{dx},\lfrac{dz}{dx}$ are both defined.

  • $\lfrac{d(x^n)}{dx} = n·x^{n-1}$ whenever $\lfrac{dx}{dt}$ is defined, for any natural $n \ge 1$.

  • $\lfrac{d(\exp(x))}{dx} = \exp(x)$ whenever $\lfrac{dx}{dt}$ is defined.

  • $\lfrac{d(\cos(x))}{dx} = -\sin(x)$ whenever $\lfrac{dx}{dt}$ is defined.

  • $\lfrac{d(\sin(x))}{dx} = \cos(x)$ whenever $\lfrac{dx}{dt}$ is defined.

  • $\lfrac{d(\ln(x))}{dx} = \lfrac1x$ whenever $x \notin \mathbb{R}_{\le 0}$ and $\lfrac{dx}{dt}$ is defined. (Here taking principal branch for $\ln$.)


In particular, we can see how these apply to your examples:

  • $\lfrac{d(\sin(3it))}{dt} = \lfrac{d(\sin(3it))}{d(3it)}·\lfrac{d(3it)}{dt} = \cos(3it)·3i$. Here $t$ is a real parameter.

  • $\lfrac{d(\sin(3iz))}{dz} = \cos(3iz)·3i$ where $z$ is a complex parameter too, corresponding to the fact that $( z \mapsto \sin(3iz) )$ has a complex derivative.

  • $\int \exp(i2t) dt = \lfrac1{2i}·\exp(i2t) + c$ for some constant $c$, because $\lfrac{d(\lfrac1{2i}·\exp(i2t))}{dt} = \exp(i2t)$ everywhere.

  • If $z = x+y·i$ where $x,y$ are real variables varying with real parameter $t$, then by the above properties $\lfrac{dz}{dt} = \lfrac{dx}{dt}+\lfrac{dy}{dt}·i$. Even if the expression for $x$ uses complex numbers, you can use the above properties to compute $\lfrac{dx}{dt}$. For example if $x = \exp(it)+\exp(-it)$, which is always real, then $\lfrac{dx}{dt} = i\exp(it)-i\exp(-it)$.

$\endgroup$
  • $\begingroup$ $\exp,\cos,\sin$ here are defined by their Taylor series, which can be motivated by wanting a solution to the differential equation $\lfrac{dy}{dx} = y$ as in this post, and which are defined on the whole complex plane. $\endgroup$ – user21820 Sep 23 '17 at 11:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.