Question Number 196265 by KRIMO last updated on 21/Aug/23
Answered by a.lgnaoui last updated on 21/Aug/23
$$\mathrm{posons}\:\mathrm{z}=\mathrm{a}+\mathrm{ib}\:\Leftrightarrow\mathrm{z}=\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \:}\:\left(\frac{\mathrm{a}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}+\mathrm{i}\frac{\mathrm{b}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}\right) \\ $$$$\mathrm{avec}\:\:\frac{\mathrm{a}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}=\mathrm{cos}\:\alpha\:\:\:\:\:\frac{\mathrm{b}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}=\mathrm{sin}\:\alpha \\ $$$$\mathrm{z}=\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right)\left(\mathrm{1}+\mathrm{i}\right) \\ $$$$\mathrm{ici}\:\:\mathrm{a}=\mathrm{b}=\mathrm{1}\:\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\:=\sqrt{\mathrm{2}} \\ $$$$\mathrm{z}=\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{i}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}\:\right)\:\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\mathrm{isin}\:\frac{\:\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$
Answered by Rodier97 last updated on 22/Aug/23
![posons Z=(1+(√3))+i(1+(√3)) Alors ∣Z∣=(√((1+(√3))^2 +(1+(√3))^2 ))=(√(2(1+(√3))^2 )) ∣Z∣=(√2)(1+(√3)) Et: { ((cosθ=((1+(√3))/( (√2)(1+(√(3))))) =((√2)/2))),((sinθ=((1+(√3) )/( (√2)(1+(√3))))=((√2)/2))) :} ⇒ θ=(π/4) D′ou^ Z=(√2)(1+(√3))[cos(π/4)+i sin(π/4)]](https://www.tinkutara.com/question/Q196269.png)
$$ \\ $$$$\:\mathrm{posons}\:\mathrm{Z}=\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)+\boldsymbol{\mathrm{i}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$$ \\ $$$$\mathrm{Alors}\:\mid\mathrm{Z}\mid=\sqrt{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{Z}\mid=\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{Et}: \\ $$$$\begin{cases}{\mathrm{cos}\theta=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\left.\mathrm{3}\right)}\right.}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{sin}\theta=\frac{\mathrm{1}+\sqrt{\mathrm{3}}\:}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{cases}\:\Rightarrow\:\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{D}'\mathrm{o}\grave {\mathrm{u}}\:\:\:\mathrm{Z}=\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\left[\mathrm{cos}\frac{\pi}{\mathrm{4}}+\boldsymbol{\mathrm{i}}\:\mathrm{sin}\frac{\pi}{\mathrm{4}}\right] \\ $$