More On Euler's Identity $e^{\ a+ib}$

Euler's identity, $e^{i\theta} = cos(\theta) + i sin(\theta)$ is useful in various ways. It is discussed with a circular plot of it on the Argand plane, which you can see at my $e^{\ i\theta}$ web page.

In addition, given that $e^{a+b} = e^{a}\ e^{b}$, we often use $e^{\ a+ib}$. Some uses of this are shown here, including a plot of it on the Argand plane.

Often we encounter functions involving $e^{at}sin(bt)$ or $e^{at}cos(bt)$, and want to differentiate or (and this is harder) integrate them. Applications include Fourier series, transforms, Laplace transforms, etc.

We show here that using Euler's identity can help - and in fact one can often process both functions in one go. Complex numbers make life simpler!

Useful for differentiation

If want, say, the differential of $e^{at}sin(bt)$, the concept is to find the differential of $e^{(a+ib)t}$ which gives the differential of both $e^{at}cos(bt)$ and $e^{at}sin(bt)$, the real part being that of $e^{at}cos(bt)$ the imaginary part that for $e^{at}sin(bt)$. One then equates the real and imaginary parts, and so finds two differentials at once.

$e^{(a+ib)t} = e^{at} \ e^{ibt} = e^{at}cos(bt) + i e^{at}sin(bt)$

$\frac{d \ e^{(a+ib)t}}{dt} = (a+ib)e^{(a+ib)t} = (a+ib)e^{at}(cos(bt) + i sin(bt))$

       $= e^{at}(a cos(bt) - b sin(bt)) + i e^{at}(b cos(bt) + a sin(bt))$

Real part $\frac{d e^{at}cos(bt)}{dt} = e^{at}(a cos(bt) - b sin(bt))$

Imag part $\frac{d e^{at}sin(bt)}{dt} = e^{at}(b cos(bt) + a sin(bt))$

This works well.

On $e^{\ (a +ib)\ \theta}$ on the Argand plane

This involves sinusoids whose amplitude varies : typically a is negative, so the amplitude decays. You can change a and b and see the result. The actual plot is $e^{a\theta} e^{i b 2 \pi \theta}$ to correspond with rotations being multiples of $2\pi$ radians.

a (-5..0) b (0..5)

Even more so for Integration

Similarly for integral of say $e^{at}sin(bt)$, find the integral of $e^{(a+ib)t}$ and use the imaginary result

$e^{(a+ib)t} = e^{at} \ e^{ibt} = e^{at}cos(bt) + i e^{at}sin(bt)$

$\int{}{} e^{(a+ib)t} dt = \frac{e^{(a+ib)t}}{a+ib} = \frac{(a-ib)e^{(a+ib)t}}{(a-ib)(a+ib)}$

       $= \frac{(a-ib)e^{at}(cos(bt) + i sin(bt))}{a^2+b^2}$

       $= \frac{e^{at}(a cos(bt) + b sin(bt)) + i e^{at}(-b cos(bt) + a sin(bt))}{a^2+b^2}$

Real part $\int{}{}e^{at}cos(bt) dt = \frac{e^{at}(a cos(bt) + b sin(bt))}{a^2+b^2}$

Imag part $\int{}{}e^{at}sin(bt) dt = \frac{e^{at}(-b cos(bt) + a sin(bt))}{a^2+b^2}$

This is simpler than using integration by parts!