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Question-206788




Question Number 206788 by SANOGO last updated on 25/Apr/24
Answered by A5T last updated on 25/Apr/24
1. ((x/y))^(−) =(x^− /y^− )⇒((1/z))^(�) =(1^− /z^− )=(1/z^− )  2. z=a+bi;z^− =a−bi;   z=z^− ⇒a+bi=a−bi⇒b=0⇒z∈R  3. ∣z∣+∣w∣≥∣z+w∣  ⇒∣z−w∣+∣w∣≥∣z−w+w∣=∣z∣  ⇒∣z∣−∣w∣≤∣z−w∣  4. z=a+bi⇒∣z∣=(√(a^2 +b^2 ))  Re(z)=a;Im(z)=bi  ∣Re(z)∣+∣Im(z)∣≥∣Re(z)+Im(z)∣  ⇒∣Re(z)∣+∣Im(z)∣≥∣a+bi∣=∣z∣...(i)  (√2)((√(a^2 +b^2 )))≥∣a∣+∣b∣=(√a^2 )+(√b^2 )=∣Re(z)∣+∣Im(z)∣...(ii)  (i)&(ii)⇒ ∣z∣≤∣Re(z)∣+∣Im(z)∣≤(√2)∣z∣
$$\mathrm{1}.\:\overline {\left(\frac{{x}}{{y}}\right)}=\frac{\overset{−} {{x}}}{\overset{−} {{y}}}\Rightarrow\overset{} {\left(\frac{\mathrm{1}}{{z}}\right)}=\frac{\overset{−} {\mathrm{1}}}{\overset{−} {{z}}}=\frac{\mathrm{1}}{\overset{−} {{z}}} \\ $$$$\mathrm{2}.\:{z}={a}+{bi};\overset{−} {{z}}={a}−{bi};\: \\ $$$${z}=\overset{−} {{z}}\Rightarrow{a}+{bi}={a}−{bi}\Rightarrow{b}=\mathrm{0}\Rightarrow{z}\in\mathbb{R} \\ $$$$\mathrm{3}.\:\mid{z}\mid+\mid{w}\mid\geqslant\mid{z}+{w}\mid \\ $$$$\Rightarrow\mid{z}−{w}\mid+\mid{w}\mid\geqslant\mid{z}−{w}+{w}\mid=\mid{z}\mid \\ $$$$\Rightarrow\mid{z}\mid−\mid{w}\mid\leqslant\mid{z}−{w}\mid \\ $$$$\mathrm{4}.\:{z}={a}+{bi}\Rightarrow\mid{z}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${Re}\left({z}\right)={a};{Im}\left({z}\right)={bi} \\ $$$$\mid{Re}\left({z}\right)\mid+\mid{Im}\left({z}\right)\mid\geqslant\mid{Re}\left({z}\right)+{Im}\left({z}\right)\mid \\ $$$$\Rightarrow\mid{Re}\left({z}\right)\mid+\mid{Im}\left({z}\right)\mid\geqslant\mid{a}+{bi}\mid=\mid{z}\mid…\left({i}\right) \\ $$$$\sqrt{\mathrm{2}}\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)\geqslant\mid{a}\mid+\mid{b}\mid=\sqrt{{a}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} }=\mid{Re}\left({z}\right)\mid+\mid{Im}\left({z}\right)\mid…\left({ii}\right) \\ $$$$\left({i}\right)\&\left({ii}\right)\Rightarrow\:\mid{z}\mid\leqslant\mid{Re}\left({z}\right)\mid+\mid{Im}\left({z}\right)\mid\leqslant\sqrt{\mathrm{2}}\mid{z}\mid \\ $$

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