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Question Number 193414 by mokys last updated on 13/Jun/23

Answered by deleteduser1 last updated on 13/Jun/23

$$Zand{Z}_{2}intheclosedunitdisk\Rightarrow \mid Z\mid ,\mid Z_2\mid ⩽1$$$$\mid Z_1-Z_2\mid ⩾1\Rightarrow (Z_1-Z_2)({\stackrel{-}{Z}}_1-{\stackrel{-}{Z}}_2)⩾1$$$$\Rightarrow \mid Z_1{\mid }^2+\mid Z_2{\mid }^2⩾1+(Z{\stackrel{-}{Z}}_2+Z_2{\stackrel{-}{Z}}_1)(since{Z}_i{\stackrel{-}{Z}}_i=\mid Z_i{\mid }^2)$$$$\Rightarrow \mid Z_1{\mid }^2+\mid Z_2{\mid }^2+Z{\stackrel{-}{Z}}_2+Z_2\stackrel{-}{Z}⩽2(\mid Z{\mid }^2+\mid Z_2{\mid }^2)-1⩽2\left(2\right)-1=3$$$$\Rightarrow \mid Z_1+Z_2{\mid }^2⩽3\Rightarrow \mid Z_1+Z_2\mid ⩽\sqrt{3}$$

Answered by witcher3 last updated on 13/Jun/23

$$\mid {\mathrm{z}}_1\mid ,\mid {\mathrm{z}}_2\mid ⩽1$$$${\mathrm{z}}_1={\mathrm{pe}}^{\mathrm{ia}}$$$${\mathrm{z}}_2={\mathrm{p}}^{\prime }{\mathrm{e}}^{\mathrm{ib}}$$$$\mid {\mathrm{z}}_1-{\mathrm{z}}_2\mid =\mid (\mathrm{pcos}\left(\mathrm{a}\right)-{\mathrm{p}}^{\prime }\cos \left(\mathrm{b}\right)+\mathrm{i}(\mathrm{psin}\left(\mathrm{a}\right)-{\mathbf{p}}^{\prime }\sin \left(\mathrm{b}\right)\mid ⩾1$$$$\Rightarrow {\mathrm{p}}^2+{\mathrm{p}}^{\prime 2}-{2\mathrm{pp}}^{\prime }\cos (\mathrm{a}-\mathrm{b})⩾1$$$${2\mathrm{pp}}^{\prime }\cos (\mathrm{a}-\mathrm{b})⩽{\mathrm{p}}^2+{\mathrm{p}}^{\prime 2}-1$$$$\mid {\mathrm{z}}_1+{\mathrm{z}}_2{\mid }^2={\mathrm{p}}^2+{\mathrm{p}}^{\prime 2}+{2\mathrm{pp}}^{\prime }\cos (\mathrm{a}-\mathrm{b})$$$$⩽{2\mathrm{p}}^2+{2\mathrm{p}}^{\prime 2}-1$$$$\mathrm{p}=\mid {\mathrm{z}}_1\mid <1,{\mathrm{p}}^{\prime }=\mid {\mathrm{z}}_2\mid <1\Rightarrow$$$$\mid {\mathrm{z}}_1+{\mathrm{z}}_2{\mid }^2⩽2+2-1=3\Rightarrow \mid {\mathrm{z}}_1+{\mathrm{z}}_2\mid ⩽\sqrt{3}$$

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