# Complex valued function of real variable¶

Let $$f:I \subseteq \mathbb{R} \to \mathbb{C}$$, such that, $$f\in \mathcal{C}^1$$, then $$\(f(x)=g(x) + ih(x)\quad x\in\mathbb{R}$$\)

Ex: $$f(t) = e^{it}$$, $$t\in\mathbb{R}$$

Ex: $$f(t)= (t+1) +(3t-1)i$$ Ex: $$f(t)=t^2 + t^3 i$$

## 1. Differentiation¶

Differentiation of such function is done term by term as follows $$f'(t)=g'(t) + ih'(t)$$

Ex:

\begin{align} f(t) &= e^{it} &=& \cos t + i \cdot \sin t \\ f'(t) &= ie^{it} &=& -\sin t + i \cos t \\ &&=& i(\cos t + i \cdot \sin t) \\ &&=& ie^{it} \\ \end{align}

We can see that, both ways we got the same anser.

## 2. Integration¶

Integration of such function is done term by term as follows, $$\int f(t) = \int g(t) dt + i \int h(t) dt$$

Ex:

\begin{align} f(t) &= t^2 + it^3 \\ \int f(t) dt &= \int t^2 dt + i \int t^3 dt \\ &= \frac{ t^3 }{ 3 } + i \frac{ t^4 }{ 4 } + C \end{align}

## 3. Legth¶

The length of the curve is conputed as the following formula

\begin{align} (ds)^2 &= (dx)^2 + (dy)^2\\ ds &= \sqrt{(dx)^2 + (dy)^2} \\ &= \sqrt{ 1 + \left(\frac{ dy }{ dx }\right)^2} dx\\ \end{align}

Integrating we get length of the curve as $$\int_a^b ds=\int_a^b\sqrt{ 1 + \left(\frac{ dy }{ dx }\right)^2}dx$$

Credits: Pinku Kumar, Pranav Kumar