Question Number 157815 by mkam last updated on 28/Oct/21
Commented by mkam last updated on 28/Oct/21
$${where}\:{z}={re}^{{i}\theta} \\ $$
Commented by mkam last updated on 28/Oct/21
$$\boldsymbol{{Solution}}: \\ $$$$ \\ $$$$\boldsymbol{{Re}}\left(\boldsymbol{{Z}}_{\mathrm{1}} .\overset{−} {\boldsymbol{{Z}}}_{\mathrm{2}} \right)\:=\:\boldsymbol{{Re}}\left(\boldsymbol{{r}}_{\mathrm{1}} .\:\boldsymbol{{e}}^{\boldsymbol{{i}\theta}_{\mathrm{1}} } \:.\boldsymbol{{r}}_{\mathrm{2}} .\boldsymbol{{e}}^{−\boldsymbol{{i}\theta}_{\mathrm{2}} } \right)\:=\:\mid\:\boldsymbol{{r}}_{\mathrm{1}} \boldsymbol{{e}}^{\boldsymbol{{i}\theta}_{\mathrm{1}} } \mid\:.\:\mid\:\boldsymbol{{r}}_{\mathrm{2}} \:.\boldsymbol{{e}}^{−\boldsymbol{{i}\theta}_{\mathrm{2}} } \:\mid \\ $$$$ \\ $$$$=\:\mid\:\boldsymbol{{r}}_{\mathrm{1}} .\boldsymbol{{r}}_{\mathrm{2}} \:.\:\boldsymbol{{e}}^{\boldsymbol{{i}}\left(\boldsymbol{\theta}_{\mathrm{1}} −\boldsymbol{\theta}_{\mathrm{2}} \right)} \:\mid\:=\mid\:\boldsymbol{{r}}_{\mathrm{1}} .\boldsymbol{{r}}_{\mathrm{2}} \mid\:.\:\mid\:\boldsymbol{{e}}^{\boldsymbol{{i}}\left(\boldsymbol{\theta}_{\mathrm{1}} −\boldsymbol{\theta}_{\mathrm{2}} \right)} \mid \\ $$$$ \\ $$$$=\:\mid\boldsymbol{{r}}_{\mathrm{1}} \mid\:.\:\mid\boldsymbol{{r}}_{\mathrm{2}} \mid\:\boldsymbol{{cos}}\left(\boldsymbol{\theta}_{\mathrm{1}} +\boldsymbol{\theta}_{\mathrm{2}} \right) \\ $$