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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

Last modified on: 2023-01-04 23:20:05