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prove-Log-z-1-z-2-Log-z-1-Log-z-2-if-pi-lt-Argz-1-Argz-2-lt-pi-hwo-can-solve-this-




Question Number 132826 by mohammad17 last updated on 16/Feb/21
prove Log(z_1 z_2 )=Log(z_1 )+Log(z_2 )  if −π<Argz_1 +Argz_2 <π  hwo can solve this
$${prove}\:{Log}\left({z}_{\mathrm{1}} {z}_{\mathrm{2}} \right)={Log}\left({z}_{\mathrm{1}} \right)+{Log}\left({z}_{\mathrm{2}} \right) \\ $$$${if}\:−\pi<{Argz}_{\mathrm{1}} +{Argz}_{\mathrm{2}} <\pi \\ $$$${hwo}\:{can}\:{solve}\:{this} \\ $$
Commented by guyyy last updated on 20/Feb/21
Answered by MJS_new last updated on 17/Feb/21
z_1 =re^(iα) ∧z_2 =se^(iβ)   log z_1 z_2  =log rse^(i(α+β))  =log r +log s +i(α+β)  log z_1  +log z_2  =log re^(iα)  +log se^(iβ)  =       =log r +iα +log s +iβ =log r +log s +i(α+β)
$${z}_{\mathrm{1}} ={r}\mathrm{e}^{\mathrm{i}\alpha} \wedge{z}_{\mathrm{2}} ={s}\mathrm{e}^{\mathrm{i}\beta} \\ $$$$\mathrm{log}\:{z}_{\mathrm{1}} {z}_{\mathrm{2}} \:=\mathrm{log}\:{rs}\mathrm{e}^{\mathrm{i}\left(\alpha+\beta\right)} \:=\mathrm{log}\:{r}\:+\mathrm{log}\:{s}\:+\mathrm{i}\left(\alpha+\beta\right) \\ $$$$\mathrm{log}\:{z}_{\mathrm{1}} \:+\mathrm{log}\:{z}_{\mathrm{2}} \:=\mathrm{log}\:{r}\mathrm{e}^{\mathrm{i}\alpha} \:+\mathrm{log}\:{s}\mathrm{e}^{\mathrm{i}\beta} \:= \\ $$$$\:\:\:\:\:=\mathrm{log}\:{r}\:+\mathrm{i}\alpha\:+\mathrm{log}\:{s}\:+\mathrm{i}\beta\:=\mathrm{log}\:{r}\:+\mathrm{log}\:{s}\:+\mathrm{i}\left(\alpha+\beta\right) \\ $$

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