# Complex integration and differentiation for $\gamma:[a,b]\rightarrow \mathbb{C}$ [closed]

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.

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.

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

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

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 $\lfrac{dx}{dt} = i\exp(it)-i\exp(-it)$.

• $\exp,\cos,\sin$ 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. – user21820 Sep 23 '17 at 11:30