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Question Number 119159 by mathocean1 last updated on 22/Oct/20

$$Wearein\mathbb{C}.$$ $$Given{Z}_0=1;Z_{n+1}=\frac{1}{2}{Z}_n+\frac{1}{2}i$$ $$n\in \mathbb{N}.$$ $$Showthat\mathrm{\forall }n\in {\mathbb{N}}^{\ast },\mid Z_n\mid <1.$$

Answered by Olaf last updated on 22/Oct/20

$$\mathrm{Let}{U}_n=Z_n-i,n\in \mathbb{N}$$ $$\Rightarrow U_0=1-i\mathrm{and}$$ $$U_{n+1}=Z_{n+1}-i=\frac{1}{2}{Z}_n-\frac{1}{2}i=\frac{1}{2}{U}_n$$ $$\Rightarrow U_n=U_0{\left(\frac{1}{2}\right)}^n=\frac{1-i}{2^n}$$ $$\mathrm{and}{Z}_n=\frac{1-i}{2^n}+i=\frac{1}{2^n}(1+(2^n-1)i)$$ $$\mid Z_n\mid =\frac{\sqrt{1+{(2^n-1)}^2}}{2^n}<\frac{\sqrt{2^{2n}}}{2^n}=1,n\in {\mathbb{N}}^{\ast }$$ $$\mathrm{and}\mid Z_0\mid =\sqrt{2}$$

Answered by mathmax by abdo last updated on 22/Oct/20

$$\mathrm{by}\mathrm{recurrence}\mathrm{n}=1\Rightarrow {\mathrm{z}}_1=\frac{1}{2}{\mathrm{z}}_0+\frac{\mathrm{i}}{2}=\frac{1}{2}+\frac{\mathrm{i}}{2}\Rightarrow \mid {\mathrm{z}}_1\mid =\frac{1}{2}\mid 1+\mathrm{i}\mid$$ $$=\frac{\sqrt{2}}{2}<1\mathrm{relation}\mathrm{true}\mathrm{for}\mathrm{n}=1\mathrm{let}\mathrm{suppise}\mid {\mathrm{z}}_{\mathrm{n}}\mid <1$$ $$\mathrm{we}\mathrm{hsve}\mid {\mathrm{z}}_{\mathrm{n}+1}\mid =\frac{1}{2}\mid {\mathrm{z}}_{\mathrm{n}}+\mathrm{i}\mid ⩽\frac{1}{2}\mid {\mathrm{z}}_{\mathrm{n}}\mid +\frac{1}{2}<\frac{1}{2}+\frac{1}{2}=1\Rightarrow \mid {\mathrm{z}}_{\mathrm{n}+1}\mid <1$$ $$\mathrm{relation}\mathrm{is}\mathrm{true}\mathrm{at}\mathrm{term}\mathrm{n}+1$$

Commented bymathocean1 last updated on 25/Oct/20

$$\mathrm{Thank}\mathrm{you}\mathrm{sirs}.$$ $$$$

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