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Question Number 70145 by Scientist0000001 last updated on 01/Oct/19

$$provethat;arg\left(\mathit{z}1\mathit{z}2\right)=arg\left(z1\right)+arg\left(z2\right).$$$$arg\left(z1/z2\right)=arg\left(z1\right)-arg\left(z2\right).$$

Answered by MJS last updated on 01/Oct/19

$$z_1=r_1{\mathrm{e}}^{\mathrm{i}{\theta }_1};z_2=r_2{\mathrm{e}}^{\mathrm{i}{\theta }_2}$$$$z_1{z}_2=r_1{r}_2{\mathrm{e}}^{\mathrm{i}({\theta }_1+{\theta }_2)}$$$$\frac{z_1}{z_2}=\frac{r_1}{r_2}{\mathrm{e}}^{\mathrm{i}({\theta }_1-{\theta }_2)}$$$$\arg \left(r{\mathrm{e}}^{\mathrm{i}\theta }\right)=\theta$$

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