Geometric Series?

Question

Let z = a + jb be a complex number with real part a and imaginary part b. Show that the series ∑ e^(a+jb)n is a geometric series and that it converges if and only if a < 0.

Answer

z = a + jb

∑ e^(a+jb)n = ∑ e^[(a+jb)n] ( in more proper writing 🙂 )

Consider the nth term and the (n+1)th term :

nth term : e^[(a+jb)n]

(n + 1) th term : e^[(a+jb)(n+1)]

The ratio of (n+1)th term to nth term
= e^[(a+jb)(n+1)] / e^[(a+jb)n]
=e^[(a+jb)(n+1) -(a+jb)n]
=e^[ (a+jb) ]
= e^z
= a constant since the complex number z is ‘pre defined’.

In other words the series ∑ e^[(a+jb)n] is a geometric series with the ‘comstant ratio’ or ‘common ratio’ of ‘e^z’.

Now when will this series converge? For a geometric series with common ratio = r, the series will converge if and only if modulus (r) < 1.

In our case,
modulus ( r ) = modulus (e^z) = modulus [ e^[ (a+jb) ] ]
= ( e^a) modulus [ e^(jb) ]

Note that we use the following ‘ Euler’s identity’ :
e^[ jx ] = Cos(x) + j Sin(x) where j = sqrt (-1) and x is any real.
Thus,
e^[ jb] = Cos(b) + j Sin(b)

So, putting this fact in the ‘above’,
modulus ( r ) = ( e^a) modulus [ e^(jb) ]
= (e^a) modulus [ Cos(b) + j Sin(b) ]
= (e^a) [ (Cos(b))^2 + ( Sin(b))^2 ]
= (e^a) [ (Cos(b))^2 + ( Sin(b))^2 ]
= (e^a) [1 ] … trigonometric identity!
= (e^a)

Thus for convergence, modulus ( r) < 1 implies
(e^a) < 1 = e^0

Since e^x is an increasing function in real numbers x,
(e^a) < e^0
implies that a < 0 as required.

Done.