Question Number 75366 by arkanmath7@gmail.com last updated on 10/Dec/19
$${i}\:{need}\:{the}\:{sol}\:{plz} \\ $$$${expansion}\:{the}\:{maclaurin}\:{series} \\ $$$${f}\left({z}\right)=\frac{{z}}{{z}^{\mathrm{4}} \:+\:\mathrm{9}}\:=\:\frac{{z}}{\mathrm{9}}\:×\frac{\mathrm{1}}{\mathrm{1}+\frac{{z}^{\mathrm{4}} }{\mathrm{9}}}\: \\ $$
Answered by Smail last updated on 10/Dec/19
$$\frac{{z}}{{z}^{\mathrm{4}} +\mathrm{9}}=\frac{{z}}{\mathrm{9}}×\frac{\mathrm{1}}{\mathrm{1}+{z}^{\mathrm{4}} /\mathrm{9}}=\frac{{z}}{\mathrm{9}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{z}^{\mathrm{4}} }{\mathrm{9}}\right)^{{n}} \\ $$$${with}\:\:\mid{z}\mid<\sqrt[{\mathrm{4}}]{\mathrm{9}} \\ $$$${f}\left({z}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{z}^{\mathrm{4}{n}+\mathrm{1}} }{\mathrm{9}^{{n}+\mathrm{1}} }\:\: \\ $$