1.

Forme trigo de $z_1=\dfrac{\sqrt{3}+i}{1-i}$ ?

$\quad\star |\sqrt{3}+i|=\sqrt{\sqrt3^2+1^2}=2$

$\quad\star |1-i|=\sqrt{1^2+(-1)^2}=\sqrt2$

Donc

$\begin{aligned} \sqrt{3}+i&=2\left(\frac{\sqrt3}2+\frac12i\right)\\ &=2\left(\cos(\dfrac{\pi}{6})+i\sin(\dfrac{\pi}{6})\right)\\ 1-i&=\sqrt2\left( \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2} \right)\\ &=\sqrt2\left( \cos(\dfrac{-\pi}{4})+i\sin(-\dfrac{\pi}{4}) \right) \end{aligned}$

et donc

$|z_1|=\dfrac{|\sqrt3+i|}{|1-i|}=\dfrac{2}{\sqrt2}=\sqrt2\\$

$\begin{aligned} \arg(z_1)&\equiv \arg\left(\dfrac{\sqrt{3}+i}{1-i}\right) ~ [2\pi] \\ &\equiv \arg(\sqrt{3}+i)-\arg(1-i) ~ [2\pi]\\ &\equiv \dfrac{\pi}{6}+\dfrac{\pi}{4} ~~[2\pi]\\ &\equiv\dfrac{5\pi}{12}~~[2\pi] \end{aligned}$

Finalement :

$\begin{aligned} z_1&=|z_1|(\cos(\theta)+i\sin(\theta))\\ &=\sqrt{2}\left(\cos(\dfrac{5\pi}{12})+i\sin(\dfrac{5\pi}{12})\right) \end{aligned}$

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Forme trigo de $z_2=(1+i)^3$ ?

$\begin{aligned} |z_2|&=|(1+i)^3|=|(1+i)|^3\\ &=\left(\sqrt{1^2+1^2}\right)^3\\ &=\sqrt{2}^3=2\sqrt{2} \end{aligned}$

$\begin{aligned} \arg(z_2)&=\arg((1+i)^3) \\ &\equiv 3\arg(1+i) ~~[2\pi] \end{aligned}$

Cherchons $\arg(1+i)$

$\begin{aligned} 1+i&=\sqrt{2}(\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})\\ &=\sqrt{2}(\cos(\dfrac{\pi}{4})+i\sin(\dfrac{\pi}{4})) \end{aligned}$

Donc $\arg(1+i)\equiv \dfrac{\pi}{4} ~~[2\pi]$

$\arg(z_2) \equiv \dfrac{3\pi}{4} ~~[2\pi]$

Finalement $\begin{aligned} z_2 &=|z_2|(\cos(\theta)+i\sin(\theta))\\ &=2\sqrt{2}(\cos(\dfrac{3\pi}{4})+i\sin(\dfrac{3\pi}{4})) \end{aligned}$

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2/a/ Forme trigo de $z=1+i$ ?

$\begin{aligned} 1+i&=\sqrt{2}(\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})\\ &=\sqrt{2}(\cos(\dfrac{\pi}{4})+i\sin(\dfrac{\pi}{4})) \end{aligned}$

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Forme trigo de $z'=\sqrt{3}-i$ ?

$|z'|=|\sqrt{3}-i|=\sqrt{\sqrt{3}^2+(-1)^2}=2$

$\begin{aligned} z'&=2\left(\dfrac{\sqrt{3}}{2}-i\dfrac{1}{2}\right)\\ &=2\left(\cos(\dfrac{\pi}{6})-i\sin(\dfrac{\pi}{6})\right)\\ &=2\left(\cos(-\dfrac{\pi}{6})+i\sin(-\dfrac{\pi}{6})\right) \end{aligned}$

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b/ Forme trigo de $z.z'$ ?

$z.z'=|z.z'|(\cos(\theta)+i\sin(\theta))$ avec $\theta \equiv \arg(z.z') ~~[2\pi]$

On a $|z.z'|=|z|.|z'|=\sqrt{2}\times2=2\sqrt{2}$

et $\arg(z.z') \equiv \arg(z)+\arg(z') ~~[2\pi]$

$\arg(z.z') \equiv \dfrac{\pi}{4}+(-\dfrac{\pi}{6}) ~~[2\pi]$

Donc $\arg(z.z') \equiv \dfrac{\pi}{12} ~~[2\pi]$

$\begin{aligned} z.z'&=|z.z'|(\cos(\theta)+i\sin(\theta))\\ &=2\sqrt{2}\left(\cos(\dfrac{\pi}{12})+i\sin(\dfrac{\pi}{12})\right) \end{aligned}$

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Forme algébrique de $z.z'$

$\begin{aligned} z.z'&=(1+i)(\sqrt{3}-i)\\ &=\sqrt{3}-i+i\sqrt{3}+1\\ &=\sqrt{3}+1+i(\sqrt{3}-1) \end{aligned}$

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c/

On a

et

Donc :