Exercices sur la forme trigonométrique des nombres complexes

1. $z_A = 4 \cdot \text{cis} \left(-\frac{\pi}{2}\right) = -4\mathbf{i}$
2. $z_B = 8 \cdot \text{cis} \left(\frac{2\pi}{3}\right) = -4+4\mathbf{i}\sqrt{3} $
3. $z_C = 4 \cdot \text{cis} \left(\frac{\pi}{4}\right) = 2\sqrt{2}+2\mathbf{i}\sqrt{2}$
4. $z_D = 9 \cdot \text{cis} \left(\frac{5\pi}{6}\right) = -\frac{9\sqrt{3}}{2}+\frac{9}{2}\mathbf{i}$
5. $z_E = 6 \cdot \text{cis} \left(-\frac{\pi}{6}\right) = \frac{3\sqrt{3}}{2} - 3 \mathbf{i}$
6. $z_F = 12 \cdot \text{cis} \left(\pi\right) = -12 \; (-12+0\cdot \mathbf{i})$

1. $z=\text{cis}(0)$
2. $z=5\cdot \text{cis}\left(\pi\right)$
3. $z=\text{cis}\left(\frac{\pi}{2}\right)$
4. $z=2\cdot \text{cis}\left(\frac{3\pi}{2}\right)$ ou $z=2\cdot \text{cis}\left(-\frac{\pi}{2}\right)$
5. $z=1-\mathbf{i} = \sqrt{2} \cdot \text{cis}\left(\frac{7\pi}{4}\right)$
6. $z=\mathbf{i}\sqrt{3}+1 = 2\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \mathbf{i}\right) = 2\cdot \text{cis} \left(\frac{\pi}{3}\right)$
7. $z=\mathbf{i}\sqrt{3}-1 = \ldots$
8. $z=\sqrt{2}\left(1+\mathbf{i}\right) = \ldots$
9. $z=\mathbf{i}^3 = \ldots$
10. $z=1+\mathbf{i}\cdot\tan(\beta) = \ldots$

1. FT : $\left(2\,\text{cis} \left ( \dfrac{\pi}{4}\right ) \right)\cdot \left(\dfrac13\;\text{cis} \left ( \dfrac{\pi}{3}\right ) \right) = \dfrac23\; \text{cis} \left ( \dfrac{7\pi}{12}\right ) $
   FA : $\left(2\,\text{cis} \left ( \dfrac{\pi}{4}\right ) \right)\cdot \left(\dfrac13\;\text{cis} \left ( \dfrac{\pi}{3}\right ) \right) = \sqrt{2} \left(1+\mathbf{i}\right)\dfrac13 \left(\dfrac12+\dfrac{\sqrt{3}}{2} \mathbf{i}\right) = \dfrac{\sqrt{2}}{3}\left(\left(1-\sqrt{3}\right)+\left(1+\sqrt{3}\right)\mathbf{i}\right)$
   Note : on peut en déduire les valeurs exactes de $\cos\left(\dfrac{7\pi}{12}\right)$ et $\sin\left(\dfrac{7\pi}{12}\right)$
2. FT : $\left(10\;\text{cis} \left ( 100^\circ\right ) \right)\cdot \left(\text{cis} \left ( 140^\circ\right ) \right) = 10\;\text{cis} \left ( 240^\circ\right ) $
   FA : $10\;\text{cis} \left ( 240^\circ \right ) = 10\left(\cos \left ( 240^\circ\right ) + \mathbf{i} \sin \left ( 240^\circ\right ) \right) = -5-5\mathbf{i}\sqrt{3}$
3. FT : $\dfrac{6\;\text{cis} \left ( 170^\circ\right ) }{3\;\text{cis} \left ( 50^\circ\right ) } = 2\;\text{cis} \left ( 120^\circ\right ) $
   FA : $\dfrac{6\;\text{cis} \left ( 170^\circ\right ) }{3\;\text{cis} \left ( 50^\circ\right ) } = 2\;\left(\cos{\left ( 120^\circ\right ) }+\mathbf{i} \sin{\left ( 120^\circ\right ) }\right) = -1+\mathbf{i}\sqrt{3}$
4. FT : $\dfrac{1}{2\;\text{cis} \left ( \dfrac{\pi}{6}\right ) } =\dfrac{\text{cis} \left ( -\dfrac{\pi}{6}\right ) }{2\;\text{cis} \left ( \dfrac{\pi}{6}\right ) \text{cis} \left ( -\dfrac{\pi}{6}\right ) } =\dfrac{\text{cis} \left ( -\dfrac{\pi}{6}\right ) }{2\;\text{cis} \left ( 0\right ) } = \dfrac12\; \text{cis} \left ( -\dfrac{\pi}{6}\right ) $
   FA : $\dfrac{1}{2\;\text{cis} \left ( \dfrac{\pi}{6}\right ) } = \dfrac12\;\left(\cos\left(-\dfrac{\pi}{6}\right)+\mathbf{i} \sin\left(-\dfrac{\pi}{6}\right)\right) = \dfrac{\sqrt{3}}{4}-\dfrac14 \mathbf{i}$
5. FT : $\dfrac{1+\mathbf{i}}{1-\mathbf{i}}=\dfrac{\sqrt{2}\cdot \text{cis} \left(\dfrac{\pi}{4}\right)}{\sqrt{2}\cdot \text{cis} \left(-\dfrac{\pi}{4}\right)}=\text{cis} \left(\dfrac{\pi}{4}-\left(-\dfrac{\pi}{4}\right)\right)=\text{cis} \left(\dfrac{\pi}{2}\right)$
   FA : $\dfrac{1+\mathbf{i}}{1-\mathbf{i}}=\text{cis} \left ( \dfrac{\pi}{2}\right ) =\mathbf{i}$
6. FT : $\mathbf{i}\; \text{cis} \left ( \dfrac{\pi}{6}\right ) = \text{cis} \left ( \dfrac{\pi}{2}\right ) \; \text{cis} \left ( \dfrac{\pi}{6} \right ) = \text{cis} \dfrac{\pi}{2}+\dfrac{\pi}{6} = \text{cis} \dfrac{2\pi}{3}$
   FA : $\mathbf{i}\; \text{cis} \left ( \dfrac{\pi}{6}\right ) = \text{cis} \left ( \dfrac{2\pi}{3}\right ) = \cos{\dfrac{2\pi}{3}} + \mathbf{i}\; \sin{\dfrac{2\pi}{3}} = -\dfrac12+\mathbf{i}\; \dfrac{\sqrt{3}}{2}$
7. FT & FA : $\dfrac{4 \;\text{cis} \left ( \dfrac{3 \pi}{4}\right ) }{2 \;\text{cis} \left ( \pi\right ) } = 2\text{cis} \left ( \dfrac{3 \pi}{4}-\pi\right ) =2\text{cis} \left ( -\dfrac{\pi}{4}\right ) = \sqrt{2}-\mathbf{i}\; \sqrt{2}$
8. FT :

   \begin{align*} \left(2 \;\text{cis} \left( \dfrac{\pi}{3}\right)\right)^2 \cdot\left(3 \;\text{cis} \left( \dfrac{\pi}{4}\right)\right)^3 &= \left(2 \;\text{cis} \left( \dfrac{\pi}{3}\right)\right)\left(2 \;\text{cis} \left( \dfrac{\pi}{3}\right)\right)\left(3 \;\text{cis} \left( \dfrac{\pi}{4}\right)\right)\left(3 \;\text{cis} \left( \dfrac{\pi}{4}\right)\right)\left(3 \;\text{cis} \left( \dfrac{\pi}{4}\right)\right)\\ &= 4 \;\text{cis}\left(\dfrac{\pi}{3}+\dfrac{\pi}{3}\right) \times 27 \;\text{cis}\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}+\dfrac{\pi}{4}\right)\\ &= 4 \;\text{cis}\left(\dfrac{2\pi}{3}\right) \times 27 \;\text{cis}\left(\dfrac{3\pi}{4}\right)\\ &= 108 \;\text{cis}\left(\dfrac{2\pi}{3}+\dfrac{3\pi}{4}\right)\\ &= 108 \;\text{cis}\left(\dfrac{17\pi}{12}\right)\\ \end{align*}

   FA :

   \begin{align*} 4 \;\text{cis}\left(\dfrac{2\pi}{3}\right) \times 27 \;\text{cis}\left(\dfrac{3\pi}{4}\right) &= 108 \left(-\dfrac12+\mathbf{i}\; \dfrac{\sqrt{3}}{2}\right)\left(-\dfrac{\sqrt{2}}2+\mathbf{i}\; \dfrac{\sqrt{2}}{2}\right)\\ &= 27 \left(-1+\mathbf{i}\;\sqrt{3}\right)\left(-\sqrt{2}+\mathbf{i}\; \sqrt{2}\right)\\ &=27\left(\left(\sqrt{2}-\sqrt{6}\right)-\left(+\sqrt{2}+\sqrt{6}\right)\mathbf{i}\right) \end{align*}

1. \(\left(1+\mathbf{i}\right)\cdot \left(\sqrt{3}-3\mathbf{i}\right) = \sqrt{2}\;\text{cis}\left(\frac{\pi}{4}\right)\; 2\sqrt{3}\;\text{cis}\left(-\frac{\pi}{3}\right) = 2\sqrt{6}\;\text{cis}\left(-\frac{\pi}{12}\right)\)
2. \(\left(1+\mathbf{i}\right) / \left(\sqrt{3}-3\mathbf{i}\right) = \frac{\sqrt{2}\;\text{cis}\left(\frac{\pi}{4}\right)}{2\sqrt{3}\;\text{cis}\left(-\frac{\pi}{3}\right)} =\frac{\sqrt{6}}{6}\;\text{cis}\left(\frac{7\pi}{12}\right)\)

Pour \( z_1 = (2+2i)^6 \), on utilise la forme trigonométrique. On a \(2+2i=2\sqrt 2\text{cis}(\pi/4)\), donc \[z_1 =(2\sqrt 2\text{cis}\left(\tfrac{\pi}{4})\right)^6 = 2^6 \cdot 2^3 \cdot \text{cis}\left(6 \cdot \tfrac{\pi}{4}\right) = 2^9 \cdot \text{cis}\left(\tfrac{3\pi}{2}\right) = 2^9 \cdot (-\textbf{i}) = -512\cdot \textbf{i}.\]

Pour \( z_2 = \left(\dfrac{1+\mathbf{i}\sqrt 3}{1-\mathbf{i}}\right)^{20} \), on calcule d'abord la forme trigonométrique du numérateur et du dénominateur. On a : \[\frac{1+\mathbf{i}\sqrt 3}{1-\mathbf{i}} = \frac{2\left(\frac12 + \textbf{i}\frac{\sqrt 3}{2}\right)}{\sqrt 2 \left(\frac{\sqrt 2}{2} - \textbf{i}\frac{\sqrt 2}{2}\right)} = \sqrt 2 \cdot \frac{\text{cis}(\pi/3)}{\text{cis}(-\pi/4)} = \sqrt2 \cdot \text{cis}\left(\tfrac{7\pi}{12}\right).\] Donc, \[ z_2 = \left(\sqrt 2 \text{cis} \left( \tfrac{7\pi}{12}\right) \right)^{20} = 2^{10} \text{cis}\left(\tfrac{140\pi}{12}\right) = 2^{10} \text{cis}\left(\tfrac{35\pi}{3}\right) = 2^{10} \text{cis}\left(\tfrac{5\pi}{3}\right) = 2^{10} (1 - \textbf{i}\sqrt 3) = 512 - 512\cdot\textbf{i}\cdot\sqrt 3.\]

Pour \( z_3 = \dfrac{(1+\mathbf{i})^{2000}}{(\mathbf{i}-\sqrt 3)^{1000}} \), on utilise également la forme trigonométrique. On a : \[(1+\mathbf{i})^{2000} = 2^{1000} \text{cis}\left( 2000\cdot \pi/4 \right) = 2^{1000} \text{cis}\left(500\cdot \pi\right) = 2^{1000},\] car \( \text{cis}(500\pi) \) revient à une rotation complète de multiples de \( 2\pi \), ce qui est équivalent à 1. De même, \[(\mathbf{i}-\sqrt 3)^{1000} = 2^{1000} \text{cis} \left( 1000 \cdot \tfrac{5\pi}{6} \right) = 2^{1000} \text{cis}\left(5000\cdot\tfrac{\pi}{6}\right).\] Donc, \[z_3 = \frac{2^{1000}}{2^{1000} \text{cis} \left(5000\cdot\frac{\pi}{6}\right) } = \text{cis} \left(-5000\cdot\tfrac{\pi}{6}\right) = \text{cis} \left(\tfrac{2\pi}{3}\right) = -\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}\mathbf{i}\]

\(z_1 = 2\;\text{cis}\left(\frac{\pi}{3}\right)\) et \(z_2=\sqrt{2}\;\text{cis}\left(-\frac{\pi}{4}\right)\)

FA : \(z_3=\frac{1+\mathbf{i}\sqrt{3}}{1-\mathbf{i}} = \frac{\left(1+\mathbf{i}\sqrt{3}\right)\left(1+\mathbf{i}\right)}{2} = \frac{1-\sqrt{3}}{2}+\frac{1+\sqrt{3}}{2}\mathbf{i}\)

FT : \(z_3=\frac{2\;\text{cis}\left(\frac{\pi}{3}\right)}{\sqrt{2}\;\text{cis}\left(-\frac{\pi}{4}\right)} = \sqrt{2}\; \text{cis}\left(\frac{\pi}{3}+\frac{\pi}{4}\right) = \sqrt{2}\; \text{cis}\left(\frac{7\pi}{12}\right)\)

On déduit du point précédent : \(\sqrt{2}\; \text{cis}\left(\frac{7\pi}{12}\right) = \frac{1-\sqrt{3}}{2}+\frac{1+\sqrt{3}}{2}\mathbf{i}\)

D'où \(\text{cis}\left(\frac{7\pi}{12}\right) = \frac{1}{\sqrt{2}}\;\left(\frac{1-\sqrt{3}}{2}+\frac{1+\sqrt{3}}{2}\mathbf{i}\right)\)

Cette égalité se réécrit : \(\cos{\left(\frac{7\pi}{12}\right)}+\mathbf{i}\; \sin{\left(\frac{7\pi}{12}\right)} = \frac{1-\sqrt{3}}{2\sqrt{2}}+\frac{1+\sqrt{3}}{2\sqrt{2}}\mathbf{i}\)

Conclusion : \(\cos{\left(\frac{7\pi}{12}\right)} = \frac{1-\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{2}-\sqrt{6}}{4}\) et \(\sin{\left(\frac{7\pi}{12}\right)} = \frac{1+\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{2}+\sqrt{6}}{4}\)

1) $z_{1}=2\;\mathrm{cis}\left(\frac{\pi}{6}\right)$ et $w=\frac{\sqrt{2}}{16}\;\mathrm{cis}\left(\frac{11\pi}{12}\right)$

2) $\left(z_{2}\right)^{3}=-16-16\mathbf{i}$ et $w = -\frac{\sqrt{3}+1}{32}+\frac{\sqrt{3}-1}{32}\mathbf{i}$

3) $\frac{\sqrt{2}}{16}\;\mathrm{cis}\left(\frac{11\pi}{12}\right) = -\frac{\sqrt{3}+1}{32}+\frac{\sqrt{3}-1}{32}\mathbf{i}$

$\mathrm{cis}\left(\frac{11\pi}{12}\right) = -\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{32}\mathbf{i}$

$\implies \cos \frac{11 \pi}{12}= -\frac{\sqrt{6}+\sqrt{2}}{4}$ et $\sin \frac{11 \pi}{12}= \frac{\sqrt{6}-\sqrt{2}}{32}$.

4) $z^{3}-z_{2}=0 \iff z^3 = 2 \sqrt{2} \operatorname{cis}\left(-\frac{\pi}{4}\right)$

on recherche donc les racines cubiques de $2 \sqrt{2} \operatorname{cis}\left(-\frac{\pi}{4}\right)$ :

$w_k = \sqrt[3]{2 \sqrt{2}} \operatorname{cis}\left(\frac{(8k-1)\pi}{12}\right) = \sqrt{2} \operatorname{cis}\left(\frac{(8k-1)\pi}{12}\right)$

1. $w_0 = \sqrt{2} \operatorname{cis}\left(-\frac{\pi}{12}\right)$
2. $w_1 = \sqrt{2} \operatorname{cis}\left(\frac{7\pi}{12}\right)$
3. $w_2 = \sqrt{2} \operatorname{cis}\left(\frac{15\pi}{12}\right)$

$S_{\mathbb C} = \left\{w_0,w_1,w_2\right\}$

5) $z_{2}^{3}-\mathbf{i} \cdot \overline{z}=i^{2022} \iff -16-16\mathbf{i}-\mathbf{i} \cdot \overline{z}=-1$

$\iff -\mathbf{i} \cdot \overline{z}=15+16\mathbf{i}$

$\iff \overline{z}=-16+15\mathbf{i}$

$S_{\mathbb C} = \left\{-16-15\mathbf{i}\right\}$

$-16+16\mathbf{i} = 2^4\left(-1+\mathbf{i}\right) = 2^4\sqrt{2}\cdot\mathrm{cis}\left(\frac{3\pi}{4}\right)$ les racines 4ème sont données par : $$ \begin{array}{l} w_0 = 2\sqrt[8]{2} \cdot \operatorname{cis}\left( \frac{3\pi}{16} + 0 \cdot \frac{\pi}{2} \right) \\ w_1 = 2\sqrt[8]{2} \cdot \operatorname{cis}\left( \frac{3\pi}{16} + 1 \cdot \frac{\pi}{2} \right) \\ w_2 = 2\sqrt[8]{2} \cdot \operatorname{cis}\left( \frac{3\pi}{16} + 2 \cdot \frac{\pi}{2} \right) \\ w_3 = 2\sqrt[8]{2} \cdot \operatorname{cis}\left( \frac{3\pi}{16} + 3 \cdot \frac{\pi}{2} \right) \\ \end{array} $$ puis simplifiées : \(w_0 = 2\sqrt[8]{2}\cdot \operatorname{cis}\left(\frac{3\pi}{16}\right) \quad ; \quad w_1 = 2\sqrt[8]{2}\cdot \operatorname{cis}\left(\frac{11\pi}{16}\right) \quad ; \quad w_2 = 2\sqrt[8]{2}\cdot \operatorname{cis}\left(\frac{19\pi}{16}\right) \quad ; \quad w_3 = 2\sqrt[8]{2}\cdot \operatorname{cis}\left(\frac{27\pi}{16}\right) \quad ; \)

Soit $z=-4=4\cdot \text{cis}\left( \pi \right)$. On a $w_k = \sqrt[4]{4} \cdot \text{cis}\left( \dfrac{\pi + 2k\pi}{4} \right)$ avec $k = 0, 1, 2, 3$ \[\newcommand{\ii}{{\mathbf{i}}} \begin{cases} w_0 = \sqrt{2} \cdot \text{cis}\left( \frac{\pi}{4} \right) &= 1 + \ii \\ w_1 = \sqrt{2} \cdot \text{cis}\left( \frac{3\pi}{4} \right) &= -1 + \ii \\ w_2 = \sqrt{2} \cdot \text{cis}\left( \frac{5\pi}{4} \right) &= -1 - \ii \\ w_3 = \sqrt{2} \cdot \text{cis}\left( \frac{7\pi}{4} \right) &= 1 - \ii \end{cases} \]

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$z^3=8\operatorname{cis}\left(\frac{3\pi}{2}\right)$

$S_{\mathbb C} = \left\{ 2\operatorname{cis}\left(\frac{\pi}{2}\right) ;2\operatorname{cis}\left(\frac{7\pi}{6}\right) ;2\operatorname{cis}\left(\frac{11\pi}{6}\right) ; \right\}$

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$Z^2+(7-{\mathbf{i}})Z -8 -8\mathbf{i} = 0 \iff Z=1+\mathbf{i} = \sqrt{2} \operatorname{cis}\left(\frac{\pi}{4}\right)$ ou $Z=-8 = 8 \operatorname{cis}\left(\pi\right)$

d'où $z^3=\sqrt{2} \operatorname{cis}\left(\frac{\pi}{4}\right)$ ou $z^3=8 \operatorname{cis}\left(\pi\right)$

$S_{\mathbb C} = \left\{ \sqrt[6]{2} \operatorname{cis}\left(-\frac{7\pi}{12}\right) ; \sqrt[6]{2} \operatorname{cis}\left(\frac{3\pi}{4}\right) ; \sqrt[6]{2} \operatorname{cis}\left(\frac{\pi}{12}\right) ; -2 ; 2 \operatorname{cis}\left(\frac{\pi}{3}\right) ; 2 \operatorname{cis}\left(-\frac{\pi}{3}\right)\right\}$

a) il suffit d'appliquer la racine cinquième de l'unité

b) $z^3=T$ ; $T^2 - 2\cos\Par{\phi}\, T+1=0$ ; $\rho = 4\left(\cos^2\Par{\phi}-1\right) = 4 \, \mathbf{i}^2\, \sin^2 \phi$

$T_{1/2}=\cos\phi\pm\mathbf{i}\, \sin\phi = \textrm{cis}\left(\pm\phi\right)$

$z_k = \textrm{cis}\left(\frac{\phi}{3} + k\frac{2\pi}{3}\right)$ et $z'_k = \textrm{cis}\left(-\frac{\phi}{3} + k\frac{2\pi}{3}\right)$ pour $k\in\left\lbrace 1,2,3 \right\rbrace$ $$S=\left\lbrace z_k, z'_k \mid k\in\left\lbrace 1,2,3 \right\rbrace \right\rbrace$$