CVEL1

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$\sum_{k=0}^{100} e^{kz} = 1 + e^z + e^{2z} + ... + e^{100z} = \mathrm{sum}\,$

If $z=2n\pi i\,$ then $\mathrm{sum} = 1 + 1 + ... + 1 = 101\,$

If $z\ne 2n\pi i\,$ then multiplying the sum by $1-e^z\,$ reveals:

$\mathrm{sum}(1-e^z) = 1-e^z+e^z-e^{2z}+e^{2z}+...+e^{100z} - e^{101z} = 1-e^{101z}\,$

Therefore $\mathrm{sum} = \frac{1-e^{101z}}{1-e^z}\,$

Also, this can be obtained from the familiar formula

$\sum_{k=0}^{100} x^{k} = 1 + x + x^{2} + ... + x^{100} = \frac{1-x^{101}}{1-x}\,$

where $x = e^z\,$

Main Page : Complex Variables : Exponential and Log

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