Question Number 132826 by mohammad17 last updated on 16/Feb/21
$${prove}\:{Log}\left({z}_{\mathrm{1}} {z}_{\mathrm{2}} \right)={Log}\left({z}_{\mathrm{1}} \right)+{Log}\left({z}_{\mathrm{2}} \right) \\ $$ $${if}\:−\pi<{Argz}_{\mathrm{1}} +{Argz}_{\mathrm{2}} <\pi \\ $$ $${hwo}\:{can}\:{solve}\:{this} \\ $$
Commented byguyyy last updated on 20/Feb/21
Answered by MJS_new last updated on 17/Feb/21
$${z}_{\mathrm{1}} ={r}\mathrm{e}^{\mathrm{i}\alpha} \wedge{z}_{\mathrm{2}} ={s}\mathrm{e}^{\mathrm{i}\beta} \\ $$ $$\mathrm{log}\:{z}_{\mathrm{1}} {z}_{\mathrm{2}} \:=\mathrm{log}\:{rs}\mathrm{e}^{\mathrm{i}\left(\alpha+\beta\right)} \:=\mathrm{log}\:{r}\:+\mathrm{log}\:{s}\:+\mathrm{i}\left(\alpha+\beta\right) \\ $$ $$\mathrm{log}\:{z}_{\mathrm{1}} \:+\mathrm{log}\:{z}_{\mathrm{2}} \:=\mathrm{log}\:{r}\mathrm{e}^{\mathrm{i}\alpha} \:+\mathrm{log}\:{s}\mathrm{e}^{\mathrm{i}\beta} \:= \\ $$ $$\:\:\:\:\:=\mathrm{log}\:{r}\:+\mathrm{i}\alpha\:+\mathrm{log}\:{s}\:+\mathrm{i}\beta\:=\mathrm{log}\:{r}\:+\mathrm{log}\:{s}\:+\mathrm{i}\left(\alpha+\beta\right) \\ $$