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Question Number 206788 by SANOGO last updated on 25/Apr/24

Answered by A5T last updated on 25/Apr/24

$$1.\overline{\left(\frac{x}{y}\right)}=\frac{\stackrel{-}{x}}{\stackrel{-}{y}}\Rightarrow \stackrel{}{\left(\frac{1}{z}\right)}=\frac{\stackrel{-}{1}}{\stackrel{-}{z}}=\frac{1}{\stackrel{-}{z}}$$$$2.z=a+bi;\stackrel{-}{z}=a-bi;$$$$z=\stackrel{-}{z}\Rightarrow a+bi=a-bi\Rightarrow b=0\Rightarrow z\in \mathbb{R}$$$$3.\mid z\mid +\mid w\mid ⩾\mid z+w\mid$$$$\Rightarrow \mid z-w\mid +\mid w\mid ⩾\mid z-w+w\mid =\mid z\mid$$$$\Rightarrow \mid z\mid -\mid w\mid ⩽\mid z-w\mid$$$$4.z=a+bi\Rightarrow \mid z\mid =\sqrt{a^2+b^2}$$$$Re\left(z\right)=a;Im\left(z\right)=bi$$$$\mid Re\left(z\right)\mid +\mid Im\left(z\right)\mid ⩾\mid Re\left(z\right)+Im\left(z\right)\mid$$$$\Rightarrow \mid Re\left(z\right)\mid +\mid Im\left(z\right)\mid ⩾\mid a+bi\mid =\mid z\mid ...\left(i\right)$$$$\sqrt{2}\left(\sqrt{a^2+b^2}\right)⩾\mid a\mid +\mid b\mid =\sqrt{a^2}+\sqrt{b^2}=\mid Re\left(z\right)\mid +\mid Im\left(z\right)\mid ...\left(ii\right)$$$$\left(i\right)\mathrm{\&}\left(ii\right)\Rightarrow \mid z\mid ⩽\mid Re\left(z\right)\mid +\mid Im\left(z\right)\mid ⩽\sqrt{2}\mid z\mid$$

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