Question Number 132633 by mohammad17 last updated on 15/Feb/21
$${find}\:{all}\:{value}\:{of}\:{z}\:{its}\:{saytisfies}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}{z}−{z}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{4}{z}}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{z}^{{n}} }{\mathrm{4}^{{n}+\mathrm{2}} } \\ $$
Answered by TheSupreme last updated on 15/Feb/21
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{z}^{{n}} }{\mathrm{4}^{{n}+\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{16}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{z}}{\mathrm{4}}\right)^{{n}} =\frac{\mathrm{1}}{\mathrm{16}}\:\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{{z}}{\mathrm{4}}}\right)=\frac{\mathrm{1}}{\mathrm{16}}\left(\frac{\mathrm{4}}{\mathrm{4}−{z}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}{z}−{z}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{4}{z}}+\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{4}−{z}\right)} \\ $$$$\frac{\mathrm{1}}{{z}\left(\mathrm{4}−{z}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\mathrm{4}{z}}+\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{4}−{z}\right)} \\ $$$$\frac{\mathrm{4}\left(\mathrm{4}−{z}\right)−\left(\mathrm{4}−{z}\right)\left(\mathrm{4}−{z}^{\mathrm{2}} \right)−{z}\left(\mathrm{4}−{z}^{\mathrm{2}} \right)}{{z}\left(\mathrm{2}−{z}\right)\left(\mathrm{2}+{z}\right)\mathrm{4}\left(\mathrm{4}−{z}\right)}=\mathrm{0} \\ $$$$\mathrm{16}−\mathrm{4}{z}−\left(\mathrm{16}−\mathrm{4}{z}^{\mathrm{2}} −\mathrm{4}{z}+{z}^{\mathrm{3}} \right)−\mathrm{4}{z}+{z}^{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{4}{z}^{\mathrm{2}} −\mathrm{4}{z}=\mathrm{0} \\ $$$$\mathrm{4}{z}\left({z}−\mathrm{1}\right)=\mathrm{0} \\ $$$${z}=\mathrm{1} \\ $$$${z}=\mathrm{0}\:…\:{out}\:{of}\:{Domain} \\ $$$${z}=\mathrm{1} \\ $$$$ \\ $$