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Question Number 19313 by Tinkutara last updated on 09/Aug/17

$$\mathrm{Prove}\mathrm{that}\mid z_1\pm z_2{\mid }^2=\mid z_2{\mid }^2+\mid z_1{\mid }^2\pm$$$$2\mathrm{Re}\left(z_1{\overline{z}}_2\right)=\mid z_1{\mid }^2+\mid z_2{\mid }^2\pm 2\mathrm{Re}({\overline{z}}_1.z_2)$$

Answered by ajfour last updated on 09/Aug/17

$$\mathrm{let}{\mathrm{z}}_1={\mathrm{x}}_1+{\mathrm{iy}}_1$$$${\mathrm{z}}_2={\mathrm{x}}_2+{\mathrm{iy}}_2$$$$\mid {\mathrm{z}}_1\pm {\mathrm{z}}_2{\mid }^2={({\mathrm{x}}_1\pm {\mathrm{x}}_2)}^2+{({\mathrm{y}}_1\pm {\mathrm{y}}_2)}^2$$$$={\mathrm{x}}_1^2+{\mathrm{y}}_1^2+{\mathrm{x}}_2^2+{\mathrm{y}}_2^2\pm 2({\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{y}}_1{\mathrm{y}}_2)$$$$\mathrm{and}{\mathrm{z}}_1{\overline{\mathrm{z}}}_2=({\mathrm{x}}_1+{\mathrm{iy}}_1)({\mathrm{x}}_2-{\mathrm{iy}}_2)$$$$\mathrm{Re}\left({\mathrm{z}}_1{\overline{\mathrm{z}}}_2\right)={\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{y}}_1{\mathrm{y}}_2$$$$\mathrm{So},\mid {\mathrm{z}}_1\pm {\mathrm{z}}_2{\mid }^2=\mid {\mathrm{z}}_1{\mid }^2+\mid {\mathrm{z}}_2{\mid }^2\pm 2\mathrm{Re}\left({\mathrm{z}}_1{\overline{\mathrm{z}}}_2\right).$$

Commented by Tinkutara last updated on 09/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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