TS – Nombres Complexes 3

$|z_1| = \sqrt{1 + 3} = 2$

donc $z_1 = 2\left(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2}\ic\right) = 2\left(\cos \dfrac{2\pi}{3} + \ic \sin \dfrac{2\pi}{3}\right)$

$\quad$

$|z_2| = \sqrt{9 + 9} = 3\sqrt{2}$

donc

$\begin{align} z_2 &= 3\sqrt{2}\left(\dfrac{1}{\sqrt{2} }-\dfrac{\ic}{\sqrt{2}}\right) \\\\
&= 3\sqrt{2} \left(\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2} \ic\right) \\\\
& = 3\sqrt{2} \left( \cos \dfrac{-\pi}{4} + \ic \sin \dfrac{-\pi}{4}\right)
\end{align}$

1. $|1 + \ic| = \sqrt{1 + 1} = \sqrt{2}$
   $1 + \ic = \sqrt{2} \left( \dfrac{1}{\sqrt{2}} + \dfrac{\ic}{\sqrt{2}} \right) = \sqrt{2} \left(\cos \dfrac{\pi}{4} + \ic \sin \dfrac{\pi}{4}\right)$
   Par conséquent arg$(1 + \ic) = \dfrac{\pi}{4} \quad (2\pi)$
   $\quad$
   $|1  – \ic| = \sqrt{1 + 1} = \sqrt{2}$.
   $1 – \ic = \sqrt{2} \left( \dfrac{1}{\sqrt{2}}-\dfrac{\ic}{\sqrt{2}} \right) = \sqrt{2} \left(\cos \dfrac{-\pi}{4} + \ic \sin \dfrac{-\pi}{4}\right)$
   Par conséquent arg$(1-\ic) = -\dfrac{\pi}{4} \quad (2\pi)$
   $\quad$
   Donc $|z| = \dfrac{\sqrt{2}}{\sqrt{2}} = 1$
   Et arg$(z) = \dfrac{\pi}{4}-\dfrac{-\pi}{4} = \dfrac{\pi}{2} \quad (2\pi)$
   $\quad$
   On pouvait également déterminer la forme algébrique de $z$ (on obtient $\ic$) et ensuite déterminer le module et un argument.
   $\quad$
2. $\left| 1 + \ic\sqrt{3}\right| = 2$
   $1 + \ic \sqrt{3} = 2\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\ic \right) = 2\left( \cos \dfrac{\pi}{3} + \ic \sin \dfrac{\pi}{3}\right)$
   Par conséquent arg$\left(1 + \ic \sqrt{3}\right) = \dfrac{\pi}{3} \quad (2\pi)$.
   $\quad$
   $|1 + \ic| = \sqrt{1 + 1} = \sqrt{2}$
   $1 + \ic = \sqrt{2} \left( \dfrac{1}{\sqrt{2}} + \dfrac{\ic}{\sqrt{2}} \right) = \sqrt{2} \left(\cos \dfrac{\pi}{4} + \ic \sin \dfrac{\pi}{4}\right)$
   Par conséquent arg$(1 + \ic) = \dfrac{\pi}{4} \quad (2\pi)$
   Donc $|z| = \dfrac{2}{\sqrt{2}} = \sqrt{2}$
   Et arg$(z) = \dfrac{\pi}{3} – \dfrac{\pi}{4} = \dfrac{\pi}{12} \quad (2\pi)$.
   $\quad$
3. $-\sqrt{2} = \sqrt{2}\left(\cos \pi + \ic \sin \pi\right)$ C’est un réel négatif!
   Donc arg$\left(-\sqrt{2} \right) = \pi \quad (2\pi)$.
   $\quad$
   $|1 + \ic| = \sqrt{1 + 1} = \sqrt{2}$
   $1 + \ic = \sqrt{2} \left( \dfrac{1}{\sqrt{2}} + \dfrac{\ic}{\sqrt{2}} \right) = \sqrt{2} \left(\cos \dfrac{\pi}{4} + \ic \sin \dfrac{\pi}{4}\right)$
   Par conséquent arg$(1 + \ic) = \dfrac{\pi}{4} \quad (2\pi)$
   $\quad$
   Donc $|z| = \dfrac{\sqrt{2}}{\sqrt{2}} = 1$
   Et arg$(z) = \pi – \dfrac{\pi}{4} = \dfrac{3\pi}{4} \quad (2\pi)$.

1. $|- 1 + \ic| = \sqrt{2}$
   Donc $-1 + \ic = \sqrt{2} \left(-\dfrac{\sqrt{2}}{2} + \ic \dfrac{\sqrt{2}}{2}\right) = \sqrt{2}\left(\cos \dfrac{3\pi}{4} + \ic \sin \dfrac{3\pi}{4}\right)$.
   Donc arg$(-1 + \ic) = \dfrac{3\pi}{4} \quad (2\pi)$.
   Par conséquent arg$\left((-1 + \ic)^5\right) = 5 \times \dfrac{3\pi}{4} \quad (2\pi) = -\dfrac{\pi}{4} \quad (2\pi)$
   $\quad$
   Ainsi
   $\begin{align} (-1 + \ic)^5 &= \sqrt{2}^5\left(\cos \dfrac{-\pi}{4}+\ic \sin \dfrac{-\pi}{4}\right) \\\\
   &= 4\sqrt{2}\left(\cos \dfrac{-\pi}{4}+\ic \sin \dfrac{-\pi}{4}\right)
   \end{align}$
   $\quad$
2. $\left|\sqrt{3}-\ic \right| = 2$.
   $\sqrt{3}-\ic = 2 \left(\dfrac{\sqrt{3}}{2}-\dfrac{\ic}{2}\right) = 2\left(\cos \dfrac{-\pi}{6} + \ic \sin \dfrac{-\pi}{6}\right)$
   Donc arg$\left(\sqrt{3}-\ic\right) = -\dfrac{\pi}{6} \quad (2\pi)$.
   $\quad$
   Par conséquent arg$\left(\left(\sqrt{3}-\ic\right)^4\right) = 4 \times \dfrac{-\pi}{6} = -\dfrac{2\pi}{3} \quad (2\pi)$.
   $\quad$
   Ainsi
   $\begin{align} \left(\sqrt{3} – \ic\right)^4 &= 2^4\left(\cos \dfrac{-2\pi}{3} + \ic \sin \dfrac{-2\pi}{3} \right) \\\\
   & = 16\left(\cos \dfrac{-2\pi}{3} + \ic \sin \dfrac{-2\pi}{3} \right)
   \end{align}$
   $\quad$
3. $\left|\left(\sqrt{2}-1\right)\ic\right| = \sqrt{2}-1$ $\quad$ arg$\left(\left(\sqrt{2}-1\right)\ic\right) = \dfrac{\pi}{2}$.
   $\quad$
   $|1-\ic| = \sqrt{2}$
   $1-\ic| = \sqrt{2}\left(\dfrac{1}{\sqrt{2}}-\dfrac{\ic}{\sqrt{2}}\right) = \sqrt{2}\left(\cos \dfrac{-\pi}{4}+\ic \sin \dfrac{-\pi}{4}\right)$
   Donc arg $(1-\ic) = -\dfrac{-\pi}{4}$
   $\quad$
   Ainsi $|z| = \dfrac{\sqrt{2}-1}{\sqrt{2}}$
   Et arg$(z) = \dfrac{\pi}{2}-\dfrac{-\pi}{4} = \dfrac{3\pi}{4}$
   $\quad$
   Donc $z = \dfrac{\sqrt{2}-1}{\sqrt{2}}\left(\cos \dfrac{3\pi}{4} + \ic \sin \dfrac{3\pi}{4} \right)$

1. $\quad$
   $\begin{align} z &= \left(\sin \dfrac{\pi}{6} + \ic \cos \dfrac{\pi}{6}\right)^6 \\\\
   & = \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\ic\right)^6 \\\\
   & = \left(\cos \dfrac{\pi}{3} + \ic \sin \dfrac{\pi}{3}\right)^6
   \end{align}$
   Par conséquent arg$(z) = 6 \times \dfrac{\pi}{3} = 2\pi \quad (2\pi) = 0 \quad (2\pi)$.
   Donc $z = \cos 0 + \ic \sin 0 $.
   $\quad$
2. $\quad$
   arg$(\ic z) = \dfrac{3\pi}{4} \quad (2\pi)$
   $\Leftrightarrow$ arg$(\ic)$ + arg$(z) = \dfrac{3\pi}{4} \quad (2\pi) $
   $ \Leftrightarrow \dfrac{\pi}{2} +$ arg$(z) = \dfrac{3\pi}{4} \quad (2\pi) $
   $\Leftrightarrow $ arg$(z) = \dfrac{\pi}{4} \quad (2\pi) $.
   Donc $z = 2\left(\cos \dfrac{\pi}{4} + \ic \sin \dfrac{\pi}{4}\right)$.

1. $|z_1| = \dfrac{\sqrt{6 + 2}}{2} = \sqrt{2}$
   Donc $z_1 = \sqrt{2} \left(\dfrac{\sqrt{3}}{2}-\dfrac{\ic}{2} \right) = \sqrt{2} \left(\cos \dfrac{-\pi}{6} + \ic \sin \dfrac{-\pi}{6}\right)$
   $\quad$
   $|z_2| = \sqrt{2}$ donc $z_2 = \sqrt{2}\left(\dfrac{1}{\sqrt{2}}-\dfrac{\ic}{\sqrt{2}}\right) = \sqrt{2}\left(\cos \dfrac{-\pi}{4} + \ic \sin \dfrac{-\pi}{4}\right)$
   $\quad$
   Par conséquent arg$\left(\dfrac{z_1}{z_2}\right) = \dfrac{-\pi}{6}-\dfrac{-\pi}{4} = \dfrac{\pi}{12} \quad (2\pi)$.
   Et $\left|\dfrac{z_1}{z_2}\right| = \dfrac{\sqrt{2}}{\sqrt{2}} = 1$.
   $\quad$
   Ainsi $\dfrac{z_1}{z_2} = \cos \dfrac{\pi}{12} + \ic \sin \dfrac{\pi}{12}$
   $\quad$
2. $\quad$
   $\begin{align} \dfrac{z_1}{z_2} &= \dfrac{\dfrac{\sqrt{6} – \ic \sqrt{2}}{2}}{1 – \ic} \\\\
   & = \dfrac{\sqrt{6}-\ic \sqrt{2}}{2(1-\ic)} \times \dfrac{1 + \ic}{1 + \ic} \\\\
   & = \dfrac{\sqrt{6} + \ic \sqrt{6}-\ic \sqrt{2} + \sqrt{2}}{4} \\\\
   &= \dfrac{\sqrt{6} + \sqrt{2}}{4} + \dfrac{\sqrt{6}-\sqrt{2}}{4} \ic
   \end{align}$
   $\quad$
3. En identifiant les formes trigonométriques et algébriques de $\dfrac{z_1}{z_2}$ on obtient :
   $\cos \dfrac{\pi}{12} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ et $\sin \dfrac{\pi}{12} = \dfrac{\sqrt{6}-\sqrt{2}}{4}$

1. $\quad$
   $\begin{align}  z_1 & = 1 + \cos \dfrac{2 \alpha}{2} + \ic \sin \dfrac{2\alpha}{2} \\\\
   & = 2\cos^2 \dfrac{\alpha}{2} + 2\ic \sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} \\\\
   & = 2\cos \dfrac{\alpha}{2} \left(\cos \dfrac{\alpha}{2} + \ic \sin \dfrac{\alpha}{2}\right)
   \end{align}$
   $\quad$
2. $\alpha \in [0;\pi|$ donc $\dfrac{\alpha}{2} \in \left[0;\dfrac{\pi}{2}\right[$
   Par conséquent $\cos \dfrac{\alpha}{2} > 0$ et $\sin \dfrac{\alpha}{2} \ge 0$
   On a donc fournit la forme trigonométrique de $z_1$.
   Ainsi $\left|z_1 \right| =2\cos \dfrac{\alpha}{2}$ et arg$(z_1) = \dfrac{\alpha}{2} \quad (2\pi)$.
   $\quad$
3. $\alpha \in [\pi;2\pi|$ donc $\dfrac{\alpha}{2} \in \left[\dfrac{\pi}{2};\pi\right[$
   Par conséquent $\cos \dfrac{\alpha}{2} < 0$ et $\sin \dfrac{\alpha}{2} \ge 0$
   Ainsi, l’expression de $z_1$ n’est donc pas donnée sous sa forme trigonométrique.
   $\begin{align} z_1 &= -2\cos \dfrac{\alpha}{2} \left(-\cos \dfrac{\alpha}{2} – \ic \sin \dfrac{\alpha}{2}\right) \\\\
   &= -2\cos \dfrac{\alpha}{2} \left(\cos \left(\pi + \dfrac{\alpha}{2}\right) + \ic \sin \left(\pi + \dfrac{\alpha}{2}\right)\right)
   \end{align}$
   Donc $\left|z_1\right| = -2\cos \dfrac{\alpha}{2}$ et arg$\left(z_1\right) = \pi + \dfrac{\alpha}{2} \quad (2\pi)$

1. $z_1 = \dfrac{1+\text{i}}{2} \times 16$ $=8(1+\text{i})$
   $z_2 = \dfrac{1+\text{i}}{2}\times 8(1 + \text{i}) $ $=8i$
   $z_3 = \dfrac{1+\text{i}}{2}\times 8\text{i} = -4 + 4\text{i}$
   $~$
2. $~$
   
3. $\left| \dfrac{1+\text{i}}{2} \right| = \dfrac{\sqrt{2}}{2}$
   Donc $\dfrac{1+\text{i}}{2} = \dfrac{\sqrt{2}}{2} \left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} \text{i} \right)$ $=\dfrac{\sqrt{2}}{2} \left( \cos  \dfrac{\pi}{4} + \text{i} \sin \dfrac{\pi}{4} \right) $
   $~$
4. $\quad$
   $\begin{align} \dfrac{z_O – z_{A_1}}{z_{A_0} – z_{A_1}} & = \dfrac{-8(1+\ic)}{16 – 8(1 + \ic)} \\\\
   &=\dfrac{-8(1 + \ic)}{8(1 – \ic)} \\\\
   &= \dfrac{-(1+ \ic)^2}{2} \\\\
   & = – \ic
   \end{align}$
   $\quad$
   Ainsi $\left|\dfrac{z_O-z_{A_1}}{z_{A_0} – z_{A_1}}\right| = \dfrac{A_1 O}{A_1 A_0} = 1$
   $\quad$
   Et arg$\left(\dfrac{z_O-z_{A_1}}{z_{A_0} – z_{A_1}} \right) = \left(\overrightarrow{A_1 O},\overrightarrow{A_1 A_0}\right) = -\dfrac{\pi}{2} \quad (2\pi)$.
   Le triangle $OA_0A_1$ est donc rectangle et isocèle en $A_1$.