Répondre :
Bonjour,
[tex]z=a+ib=r(cos(\phi)+isin(\phi))[/tex]
1) [tex]z=1+2i[/tex]
[tex]r=\sqrt{a^2+b^2}=\sqrt{1^2+2^2}=\sqrt{1+4}=\sqrt{5[/tex]
[tex]\left\{\begin{matrix}cos(\phi)=\dfrac{a}{r} \\\\sin(\phi)=\dfrac{b}{r} \end{matrix}\right.\ \ \ \Longrightarrow\left\{\begin{matrix}cos(\phi)=\dfrac{1}{\sqrt{5}} \\\\sin(\phi)=\dfrac{2}{\sqrt{5}} \end{matrix}\right.\ \ \ \Longrightarrow\phi=1,1+k2\pi[/tex]
D'où [tex]z=\sqrt{5}(cos(1,1)+isin(1,1))=\sqrt{5}cis(1,1)[/tex]
2) [tex]z=1+\sqrt{3}+i[/tex]
[tex]r=\sqrt{a^2+b^2}=\sqrt{(1+\sqrt{3})^2+1^2}=\sqrt{1+2\sqrt{3}+3+1}=\sqrt{5+2\sqrt{3}}[/tex]
[tex]\left\{\begin{matrix}cos(\phi)=\dfrac{a}{r} \\\\sin(\phi)=\dfrac{b}{r} \end{matrix}\right.\ \ \ \Longrightarrow\left\{\begin{matrix}cos(\phi)=\dfrac{1+\sqrt{3}}{\sqrt{5+2\sqrt{3}}} \\\\sin(\phi)=\dfrac{1}{\sqrt{5+2\sqrt{3}}} \end{matrix}\right.\ \ \ \Longrightarrow\phi=0,35+k2\pi[/tex]
D'où [tex]z=\sqrt{5+2\sqrt{3}}(cos(0,35)+isin(0,35))=\sqrt{5+2\sqrt{3}}cis(0,35)[/tex]
[tex]z=a+ib=r(cos(\phi)+isin(\phi))[/tex]
1) [tex]z=1+2i[/tex]
[tex]r=\sqrt{a^2+b^2}=\sqrt{1^2+2^2}=\sqrt{1+4}=\sqrt{5[/tex]
[tex]\left\{\begin{matrix}cos(\phi)=\dfrac{a}{r} \\\\sin(\phi)=\dfrac{b}{r} \end{matrix}\right.\ \ \ \Longrightarrow\left\{\begin{matrix}cos(\phi)=\dfrac{1}{\sqrt{5}} \\\\sin(\phi)=\dfrac{2}{\sqrt{5}} \end{matrix}\right.\ \ \ \Longrightarrow\phi=1,1+k2\pi[/tex]
D'où [tex]z=\sqrt{5}(cos(1,1)+isin(1,1))=\sqrt{5}cis(1,1)[/tex]
2) [tex]z=1+\sqrt{3}+i[/tex]
[tex]r=\sqrt{a^2+b^2}=\sqrt{(1+\sqrt{3})^2+1^2}=\sqrt{1+2\sqrt{3}+3+1}=\sqrt{5+2\sqrt{3}}[/tex]
[tex]\left\{\begin{matrix}cos(\phi)=\dfrac{a}{r} \\\\sin(\phi)=\dfrac{b}{r} \end{matrix}\right.\ \ \ \Longrightarrow\left\{\begin{matrix}cos(\phi)=\dfrac{1+\sqrt{3}}{\sqrt{5+2\sqrt{3}}} \\\\sin(\phi)=\dfrac{1}{\sqrt{5+2\sqrt{3}}} \end{matrix}\right.\ \ \ \Longrightarrow\phi=0,35+k2\pi[/tex]
D'où [tex]z=\sqrt{5+2\sqrt{3}}(cos(0,35)+isin(0,35))=\sqrt{5+2\sqrt{3}}cis(0,35)[/tex]