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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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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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Taylor and Laurent Series - Complex Analysis

Find the Taylor Series expansion of

It is well known (can be shown by the definition) that that Making the appropriate substitutions Shifting the index of summation gives us

Find the first three terms of the Taylor Series of

Taylor Series expansion of the numerator and the denominator seperately gives us A term by term multiplication gives us Combining like terms

Find the Taylor Series expansion of

It is well known (can be shown by the definition) that that Making the appropriate substitutions Shifting the index of summation gives us

Find the first three terms of the Taylor Series of

Taylor Series expansion of the numerator and the denominator seperately gives us A term by term multiplication gives us Combining like terms

Find the Taylor Series expansion of

It is well known (can be shown by the definition) that that Making the appropriate substitutions Shifting the index of summation gives us

Find the first three terms of the Taylor Series of

Taylor Series expansion of the numerator and the denominator seperately gives us A term by term multiplication gives us Combining like terms

Find the Taylor Series expansion of

It is well known (can be shown by the definition) that that Making the appropriate substitutions Shifting the index of summation gives us

Find the first three terms of the Taylor Series of

Taylor Series expansion of the numerator and the denominator seperately gives us A term by term multiplication gives us Combining like terms

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

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

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

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

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

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

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

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