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Question Number 70145 by Scientist0000001 last updated on 01/Oct/19
prove that ; arg(z1z2)=arg(z1)+arg(z2).  arg(z1/z2)=arg(z1)−arg(z2).
$${prove}\:{that}\:;\:{arg}\left(\boldsymbol{{z}}\mathrm{1}\boldsymbol{{z}}\mathrm{2}\right)={arg}\left({z}\mathrm{1}\right)+{arg}\left({z}\mathrm{2}\right). \\ $$$${arg}\left({z}\mathrm{1}/{z}\mathrm{2}\right)={arg}\left({z}\mathrm{1}\right)−{arg}\left({z}\mathrm{2}\right). \\ $$
Answered by MJS last updated on 01/Oct/19
z_1 =r_1 e^(iθ_1 ) ; z_2 =r_2 e^(iθ_2 )   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 ))   arg (re^(iθ) )=θ
$${z}_{\mathrm{1}} ={r}_{\mathrm{1}} \mathrm{e}^{\mathrm{i}\theta_{\mathrm{1}} } ;\:{z}_{\mathrm{2}} ={r}_{\mathrm{2}} \mathrm{e}^{\mathrm{i}\theta_{\mathrm{2}} } \\ $$$${z}_{\mathrm{1}} {z}_{\mathrm{2}} ={r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{e}^{\mathrm{i}\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} \\ $$$$\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }=\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }\mathrm{e}^{\mathrm{i}\left(\theta_{\mathrm{1}} −\theta_{\mathrm{2}} \right)} \\ $$$$\mathrm{arg}\:\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)=\theta \\ $$

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