Taylor and Laurent Series - Complex Analysis

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Question

Find the Taylor Series expansion of $f(z) = z^2e^{3z}$

Answer

It is well known (can be shown by the definition) that that

$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$

Making the appropriate substitutions

$e^{3z} = \sum_{n=0}^\infty \frac{3^nz^n}{n!}$

$z^2e^{3z} = \sum_{n=0}^\infty \frac{3^nz^{n+2}}{n!}$

Shifting the index of summation gives us

$z^2e^{3z} = \sum_{n=2}^\infty \frac{3^{n-2}z^{n}}{(n-2)!}$

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Question

Find the first three terms of the Taylor Series of $\frac{e^z}{1+z} \text{ where }|z|<1$

Answer

Taylor Series expansion of the numerator and the denominator seperately gives us

$e^z = 1+z+\frac{z^2}{2} + \frac{z^3}{6} + .....$ $\frac{1}{1+z} = \frac{1}{1-(-z)} = 1 - z + z^2 - z^3 + ....$

A term by term multiplication gives us

$(1+z+\frac{1}{2}z^2+\frac{1}{6}z^3+...) + (-z-z^2-\frac{1}{2}z^3-\frac{1}{6}z^4 + ...) + (z^2+z^3+\frac{1}{2}z^4+\frac{1}{6}z^5 + ...) + (-z^3-z^4 -\frac{1}{2}z^5-\frac{1}{6}z^6 + ...)$

Combining like terms

$\frac{e^z}{1+z} = 1 + \frac{1}{2}z^2 - \frac{1}{3}z^3 + ....$

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