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Question Number 118780 by mathocean1 last updated on 19/Oct/20

z and z′ ∈ C . show that: 1. zz′^(−) =z^− ×z′^(−) 2. ((z/(z′)))^(−) =(z^− /(z′^(−) ))

$$\mathrm{z}\:\mathrm{and}\:\mathrm{z}'\:\in\:\mathbb{C}\:. \\ $$$$\mathrm{show}\:\mathrm{that}: \\ $$$$\mathrm{1}.\:\:\:\:\:\:\overline {\mathrm{zz}'}=\overset{−} {\mathrm{z}}×\overline {\mathrm{z}'} \\ $$$$\mathrm{2}.\:\:\:\:\:\:\:\overline {\left(\frac{\mathrm{z}}{\mathrm{z}'}\right)}=\frac{\overset{−} {\mathrm{z}}}{\overline {\mathrm{z}'}} \\ $$$$ \\ $$

Answered by MJS_new last updated on 19/Oct/20

z_n =r_n e^(iθ_n ) z_n ^− =r_n e^(−iθ_n ) z_1 z_2 =r_1 r_2 e^(i(θ_1 +θ_2 )) z_1 z_2 ^(−) =r_1 r_2 e^(−i(θ_1 +θ_2 )) =r_1 e^(−iθ_1 ) ×r_2 e^(−iθ_2 ) =z_1 ^(−) ×z_2 ^(−) similar for division

$${z}_{{n}} ={r}_{{n}} \mathrm{e}^{\mathrm{i}\theta_{{n}} } \\ $$$$\overset{−} {{z}}_{{n}} ={r}_{{n}} \mathrm{e}^{−\mathrm{i}\theta_{{n}} } \\ $$$${z}_{\mathrm{1}} {z}_{\mathrm{2}} ={r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{e}^{\mathrm{i}\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} \\ $$$$\overline {{z}_{\mathrm{1}} {z}_{\mathrm{2}} }={r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{e}^{−\mathrm{i}\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} ={r}_{\mathrm{1}} \mathrm{e}^{−\mathrm{i}\theta_{\mathrm{1}} } ×{r}_{\mathrm{2}} \mathrm{e}^{−\mathrm{i}\theta_{\mathrm{2}} } =\overline {{z}_{\mathrm{1}} }×\overline {{z}_{\mathrm{2}} } \\ $$$$\mathrm{similar}\:\mathrm{for}\:\mathrm{division} \\ $$

Commented by mathocean1 last updated on 19/Oct/20

thanks

$$\mathrm{thanks} \\ $$$$ \\ $$$$ \\ $$

Answered by Olaf last updated on 19/Oct/20

z = a+ib and z′ = a′+ib′ z^_ = a−ib and z′^(_) = a′−ib′ zz′ = aa′−bb′+(ab′+a′b)i zz′^(__) = aa′−bb′−(ab′+a′b)i (1) z^_ z′^(_) = (a−ib)(a′−ib′) z^_ z′^(_) = aa′−bb′−(ab′+a′b) (2) With (1) and (2) : zz′^(__) = z^_ z′^(_) Same reasoning for question 2.

$${z}\:=\:{a}+{ib}\:\mathrm{and}\:{z}'\:=\:{a}'+{ib}' \\ $$$$\overset{\_} {{z}}\:=\:{a}−{ib}\:\mathrm{and}\:\overset{\_} {{z}'}\:=\:{a}'−{ib}' \\ $$$$ \\ $$$${zz}'\:=\:{aa}'−{bb}'+\left({ab}'+{a}'{b}\right){i} \\ $$$$\overset{\_\_} {{zz}'}\:=\:{aa}'−{bb}'−\left({ab}'+{a}'{b}\right){i}\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\overset{\_} {{z}}\overset{\_} {{z}'}\:=\:\left({a}−{ib}\right)\left({a}'−{ib}'\right) \\ $$$$\overset{\_} {{z}}\overset{\_} {{z}'}\:=\:{aa}'−{bb}'−\left({ab}'+{a}'{b}\right)\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{With}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\::\:\overset{\_\_} {{zz}'}\:=\:\overset{\_} {{z}}\overset{\_} {{z}'} \\ $$$$ \\ $$$$\mathrm{Same}\:\mathrm{reasoning}\:\mathrm{for}\:\mathrm{question}\:\mathrm{2}. \\ $$

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