TITLE: Linearity of the Riemann integral with complex valued functions
QUESTION [1 upvotes]: I want to verify the following facts for Complex integrals assuming that this works for real integrals:

Let $f,g:[a,b] \to \mathbb{C}$ and $a \in \mathbb{C}$.
  (a) $\int_{a}^{b}f(t)+g(t)\,dt=\int_{a}^{b}f(t)\,dt+\int_{a}^{b}g(t)\,dt$
  (b) $a\int_{a}^{b}f(t)\,dt=\int_{a}^{b}af(t)\,dt$

(a) Taking advantage of the linearity of the integral within real numbers.
$\int_a^b f(t)+g(t)\,dt=\int_a^b u_f(t)+iv_f(t)+u_g(t)+iv_g(t)\,dt=\int_a^b u_f(t)+iv_f(t)\,dt+\int_a^b u_g(t)+iv_g(t)\,dt=\int_a^b f(t)\,dt+\int_a^b g(t)\,dt$
(b) $a\int_a^b f(t)\,dt=(x+iy)\int_a^b u(t)+iv(t)\,dt=(x+iy)\int_a^b u(t)\,dt+(x+iy)i\int_a^b v(t)\,dt=x\int_a^b u(t)\,dt+iy \int_a^b u(t)\,dt+ix\int_a^b v(t)\,dt-y\int_a^b v(t)\,dt=\int_a^b xu(t)\,dt+i\int_a^b xv(t)\,dt+i\int_a^b yu(t)\,dt-\int_a^b yv(t)\,dt=\int_a^b xu(t)\,dt-\int_a^b y v(t)\,dt+i \int_a^b x v(t)\,dt+i\int_a^b yu(t)\,dt=\int_a^b a f(t)\,dt$

REPLY [0 votes]: In math we sometimes need just to 'compute' things out to 'prove' two objects are the same.
I our case we want to prove that $\int_{a}^{b}f(t)+g(t)\,dt$ gives the same complex number as $\int_{a}^{b}f(t)\,dt+\int_{a}^{b}g(t)\,dt$
Your calculation, which uses the linearity of the real Integral is valid as we can split any complex function $f(t)$ in two parts:$$f(t)=Re(f(t))+i\,Im(f(t))=:u(t)+i\,v(t)$$where $u(t)$ and $v(t)$ are real valued functions.
Now for the second part you should maybe show that if $z,w\in \mathbb{C}$ then it holds that:$$z\cdot w=(Re(z)+i\,Im(z))\cdot(Re(w)+i\,Im(w))=(Re(z)\cdot  Re(w)-Im(z)\cdot Im(w))+i\,(Re(z)\cdot Im(w)+Im(z)\cdot Re(w))=:c\in \mathbb{C} \tag{1}$$
and then your last '$=$' also holds as:$$\int_a^b a f(t)\,dt\overset{(1)}{=}\int_a^b (Re(a)\cdot  Re(f(t))-Im(a)\cdot Im(f(t)))+i\,(Re(a)\cdot Im(f(t))+Im(a)\cdot Re(f(t)))\,dt \overset{linearity\, of\, \mathbb{R}\, Integral}{=}\int_a^b Re(a)\cdot  Re(f(t))\,dt-\int_a^b Im(a)\cdot Im(f(t))\,dt+i\,\int_a^b Re(a)\cdot Im(f(t))\,dt+i\,\int_a^b Im(a)\cdot Re(f(t))\,dt \overset{definition}{=} \int_a^b xu(t)\,dt-\int_a^b y v(t)\,dt+i \int_a^b x v(t)\,dt+i\int_a^b yu(t)\,dt$$
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