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

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.

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}

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