N°52 Page 81
Complexes - Trigonométrie
On donne :
$$\textcolor{#caa7ff}{
z_1 = 3e^{i \frac{\pi}{4}},
z_2 = 5e^{-i \frac{\pi}{6}},
z_3 = \sqrt{3}e^{i \frac{5 \pi}{6}}
}$$
Ecrire sous forme exponentielle, puis sous forme algébrique chacun des nombres complexes.
a) $\textcolor{#caa7ff}{\dfrac{z_1z_3}{z_2}}$
b) $\textcolor{#caa7ff}{\dfrac{z_1z_2}{z_3}}$
c) $\textcolor{#caa7ff}{z_2^3}$
a) Forme exponentielle :
$$\textcolor{#caa7ff}{
\dfrac{z_1z_3}{z_2} \newline =
\dfrac{3e^{i \frac{\pi}{4}} \sqrt{3}e^{i \frac{5 \pi}{6}}}{5e^{-i \frac{\pi}{6}}} \newline =
\dfrac{3\sqrt{3}e^{i \frac{\pi}{4} + i \frac{5\pi}{6} + i \frac{\pi}{6}}}{5} \newline =
\dfrac{3\sqrt{3}}{5}
e^{i(\frac{3\pi}{12} + \frac{10\pi}{12} + \frac{2\pi}{12})} \newline =
\dfrac{3\sqrt{3}}{5}
e^{i\frac{15\pi}{12}} \newline =
\boxed{
\dfrac{3\sqrt{3}}{5}
e^{i\frac{5\pi}{4}}
}
}$$
Forme algébrique :
$$\textcolor{#caa7ff}{
\dfrac{z_1z_3}{z_2} \newline =
\dfrac{3\sqrt{3}}{5}
e^{i\frac{5\pi}{4}} \newline =
\dfrac{3\sqrt{3}}{5}
(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}) \newline =
\dfrac{3\sqrt{3}}{5}
(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}) \newline =
\dfrac{3\sqrt{3}}{5}
(-\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}) \newline =
\boxed{
-\dfrac{3\sqrt{6}}{10} - i \dfrac{3\sqrt{6}}{10}
}
}$$
b) Forme exponentielle :
$$\textcolor{#caa7ff}{
\dfrac{z_1z_2}{z_3} \newline =
\dfrac{3e^{i \frac{\pi}{4}}5e^{-i \frac{\pi}{6}}}{\sqrt{3}e^{i \frac{5 \pi}{6}}} \newline =
\dfrac{15}{\sqrt{3}}
e^{i \frac{\pi}{4} - i \frac{\pi}{6} - i \frac{5\pi}{6}} \newline =
5\sqrt{3}
e^{i(\frac{3\pi}{12} - \frac{2\pi}{12} - \frac{10\pi}{12})} \newline =
\boxed{
5\sqrt{3}
e^{i \frac{-3\pi}{4}}
}
}$$
Forme algébrique :
$$\textcolor{#caa7ff}{
\dfrac{z_1z_2}{z_3} \newline =
5\sqrt{3}
e^{i \frac{-3\pi}{4}} \newline =
5\sqrt{3}
(\cos \frac{-3\pi}{4} + i \sin \frac{-3\pi}{4}) \newline =
5\sqrt{3}
(-\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}) \newline =
\boxed{
-\dfrac{5\sqrt{6}}{2} - i\dfrac{5\sqrt{6}}{2}
}
}$$
c) Forme exponentielle :
$$\textcolor{#caa7ff}{
z_2^3 \newline =
(5e^{-i \frac{\pi}{6}})^3 \newline =
\boxed{
125e^{i \frac{-\pi}{2}}
}
}$$
Forme algébrique :
$$\textcolor{#caa7ff}{
z_2^3 \newline =
125e^{i \frac{-\pi}{2}} \newline =
125
(\cos \frac{-\pi}{2} + i \sin \frac{-\pi}{2}) \newline =
\boxed{-125i}
}$$