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  • Geometric Series?

    Anonymous Christian
    Sep 5, 2014
    MATH
    Comments Off on 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.

     

anonymous-christian

Jonathan Ramachandran is founder at AnonymousChristian.org which serves as a Christian Blog discussing the Greater Hope of Biblical Christian in Christ Alone. He is also an IMO (International Math Coach) and a featuring Guitarist for many bands.

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