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Question Number 55174 by gunawan last updated on 18/Feb/19

$$\mathrm{center}\mathrm{and}\mathrm{radius}\mathrm{convergence}$$$$\mathrm{of}\mathrm{series}\underset{n=0}{\stackrel{\propto }{\mathrm{\Sigma }}}{\left(\frac{4-2i}{1+5i}\right)}^n{z}^n\mathrm{is}...$$$$$$

Commented by Abdo msup. last updated on 19/Feb/19

$$let{u}_n\left(z\right)={\left(\frac{4-2i}{1+5i}\right)}^n{z}^{n}forz\ne 0$$$$\mid \frac{u_{n+1}\left(z\right)}{u_n\left(z\right)}\mid =\mid \frac{4-2i}{1+5i}\mid \mid z\mid =\frac{\sqrt{16+4}}{\sqrt{1+25}}\mid z\mid$$$$=\sqrt{\frac{20}{26}}\mid z\mid =\sqrt{\frac{10}{13}}\mid z\mid$$$$if\sqrt{\frac{10}{13}}\mid z\mid <1\iff \mid z\mid <\sqrt{\frac{13}{10}}theserieconverges$$$$if\sqrt{\frac{10}{13}}\mid z\mid >1\iff \mid z\mid >\sqrt{\frac{13}{10}}theseriediverges$$$$radiusisR=\sqrt{\frac{13}{10}}$$

Commented by gunawan last updated on 19/Feb/19

$$\mathrm{wow}$$$$\mathrm{thanks}\mathrm{Sir}$$

Commented by maxmathsup by imad last updated on 19/Feb/19

$$youarewelcome.$$

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