Question

For the function
f(z)=1/(z²(1 + z + 2z2))
,
find the first three terms of the Laurent Series expansion of f about a = 0 that converges
in the deleted disk D'(0, δ) for some δ > 0

Answer 1

$\begin{aligned} f(z) &= \frac{1}{z^2(1 + z + 2z^2)}\\ &= \frac{1}{2z^2\left(\frac{1 + z}{2}+ z^2\right)} \\ &= \frac{1}{2z^2\left(\left(z + \frac{1}{4}\right)^2 + \frac{7}{16}\right)}\\ &= \frac{1}{2z^2\left(\left(z + \frac{1}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2\right)}\\ &= \frac{1}{2z^2\left(z + \frac{1 - j\sqrt{7}}{4}\right)\left(z + \frac{1 + j\sqrt{7}}{4}\right)}\\\\&\hspace{0.5cm}\{\textsf{where} \hspace{0.1cm}j \hspace{0.1cm}\textsf{is a complex number}\}\\ &= \frac{1}{2} \cdot \frac{4}{2jz^2\sqrt{7}} \left(\frac{1}{\left(z + \frac{1 - j\sqrt{7}}{4}\right)} - \frac{1}{\left(z + \frac{1 + j\sqrt{7}}{4}\right)}\right) \\&=\frac{-j}{z^2 \sqrt{7}} \left(\frac{4}{1 - j\sqrt{7}}\cdot\frac{1}{\frac{4}{1 - j\sqrt{7}} z + 1} - \frac{4}{1 + j\sqrt{7}}\cdot\frac{1}{\frac{4}{1 + j\sqrt{7}} z + 1}\right) \\&=\frac{-j}{z^2 \sqrt{7}} \left(\frac{4}{1 - j\sqrt{7}}\left(1 - \frac{4}{1 - j\sqrt{7}}z + \left(\frac{4}{1 - j\sqrt{7}}z\right)^2 - \left(\frac{4}{1 - j\sqrt{7}}z\right)^3 +...\right) - \right.\\&\left.\frac{4}{1 + j\sqrt{7}}\left(1 - \frac{4}{1 + j\sqrt{7}}z + \left(\frac{4}{1 + j\sqrt{7}}z\right)^2 - \left(\frac{4}{1 + j\sqrt{7}}z\right)^3 +...\right) \right) \\&=\frac{-j}{z^2 \sqrt{7}} \left(\frac{4}{1 - j\sqrt{7}} - \frac{4}{1 + j\sqrt{7}} -\right.\\& \left(\frac{4}{1 - j\sqrt{7}}\right)^2 z + \left(\frac{4}{1 - j\sqrt{7}}\right)^3 z^2 - \left(\frac{4}{1 - j\sqrt{7}}\right)^4 z^3 +...+\\&\left.\left(\frac{4}{1 + j\sqrt{7}}\right)^2 z - \left(\frac{4}{1 + j\sqrt{7}}\right)^3 z^2 + \left(\frac{4}{1 + j\sqrt{7}}\right)^4 z^3 -...\right) \\&=\frac{-j}{z^2 \sqrt{7}} \left(\frac{4(1 + j\sqrt{7} - 1 + j\sqrt{7})}{8} +\right.\\& \frac{4^2 z}{8^2} ((1 - j\sqrt{7})^2 - (1 + j\sqrt{7})^2) + \\&\left.\frac{4^3 z^2}{8^3} ((1 + j\sqrt{7})^3 - (1 - j\sqrt{7})^3) + \frac{4^4 z^3}{8^4} ((1 - j\sqrt{7})^4 - (1 + j\sqrt{7})^4) +...\right) \\&=\frac{-j}{z^2 \sqrt{7}} \left( \frac{j8\sqrt{7}}{8} + \frac{4^2 z}{8^2} (-4j\sqrt{7}) + \right.\\&\frac{4^3 z^2}{8^3} (3(j\sqrt{7}) + 3(j\sqrt{7}) + (j\sqrt{7})^3 + (j\sqrt{7})^3) +\\&\left. \frac{4^4 z^3}{8^4} (4(-j\sqrt{7}) - 4(j\sqrt{7}) + 4(-j\sqrt{7})^3 - 4(j\sqrt{7})^3) +...\right) \\&=\frac{-j}{z^2 \sqrt{7}} \left(\frac{j8\sqrt{7}}{8} + \frac{4^2 z}{8^2} (-4j\sqrt{7}) + \right.\\&\frac{4^3 z^2}{8^3} (-8j\sqrt{7}) + \frac{4^4 z^3}{8^4}(48j\sqrt{7}) +\\&\left. \frac{4^5 z^4}{8^5}(-32j\sqrt{7}) + O(z^5)\right) \\&=\frac{-j^2}{z^2} \left( 1+ \frac{z}{4} (-4) + \frac{z^2}{8} (-8) + \frac{z^3}{16}(48) + \frac{z^4}{32}(-32) + O(z^5)\right) \\&=\frac{1}{z^2} \left( 1+ \frac{z}{4} (-4) + \frac{z^2}{8} (-8) + \frac{z^3}{16}(48) + \frac{z^4}{32}(-32) + O(z^5)\right) \\&=\frac{1}{z^2} \left( 1 - z - z^2 + 3z^3 - z^4 + O(z^5)\right) \\&=\frac{1}{z^2} - \frac{1}{z} - 1 + 3z - z^2 + O(z^3) \end{aligned} \therefore \hspace{0.1cm} \textsf{The first three terms of the}\\\textsf{ Laurent series} \hspace{0.1cm}f(z) \hspace{0.1cm} \textsf{is} \\ \displaystyle f(z) = \frac{1}{z^2} - \frac{1}{z} - 1$