1. On a: \begin{align*} 1+i\sqrt 3&=2\left(\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\right)\\ 1+i\sqrt 3&=2~e^{\left(i\frac{\pi}{3}\right)}\\ \end{align*} D'autres part: \begin{align*} 1+i&=\sqrt{2}\left(\dfrac{\sqrt 2}{2}+i\dfrac{\sqrt{2}}{2}\right)\\ 1+i&=\sqrt{2}~e^{\left(i \frac{\pi}{4}\right)} \end{align*} En outre: \begin{align*} \dfrac{1+i\sqrt 3}{1+i}&=\sqrt{2}e^{\left(i\frac{\pi}{3}-\frac{\pi}{4}\right)}\\ &=\sqrt{2}e^{\left(i\frac{\pi}{12}\right)} \end{align*} Et donc: \begin{align*} z_1&=2^{\left(\frac{n}{2}\right)}~e^{i\left(\frac{n\pi}{12}\right)}\\ z_1&=2^{\left(\frac{n}{2}\right)}~\cos\left(\frac{n\pi}{12}\right)+i~2^{\left(\frac{n}{2}\right)}~\cos\left(\frac{n\pi}{12}\right) \end{align*}

  2. \begin{align*} (1+\cos(\theta) +i\sin(\theta))&=(1+e^{i\theta})\\ (1+\cos(\theta) +i\sin(\theta))&=e^{\left(\frac{\theta}{2}\right)}~(e^{-i\theta} + e^{i\theta})\\ (1+\cos(\theta) +i\sin(\theta))&=2~\cos\left(\frac{\theta}{2}\right)~e^{\frac{i\theta}{2}} \end{align*} On en déduit les formes géométrique et algébriques: \begin{align*} z_2&=2^{2n}~\cos^{2n}\left(\frac{\theta}{2}\right)~e^{in\theta}\\ z_2&=2^{2n}~\cos^{2n}\left(\frac{\theta} {2}\right)\cos(n\theta)+i~2^{2n}\cos^{2n}\left(\frac{\theta}{2}\right)\sin(n\theta)\\ \end{align*}
  3. On a:
    \begin{align*} (1-i)^n-(\sqrt{2})^n&=(\sqrt{2})^n\left(\left(\dfrac{1}{\sqrt{2}}-i\dfrac{1}{\sqrt{2}}\right)^n-1\right)\\ (1-i)^n-(\sqrt{2})^n&=2^{\frac{n}{2}}(e^{-i\frac{n\pi}{4}}-1) \end{align*} De la mĂȘme maniĂšre on montre que: $$(1+i)^n-(\sqrt{2})^n=2^{\frac{n}{2}}(e^{i\frac{n\pi}{4}}-1)$$ Par la suite: \begin{align*} z_3&=\dfrac{e^{-i\left(\frac{n\pi}{4}\right)}-1}{e^{i\left(\frac{n\pi}{4}\right)}-1}\\ z_3&=\dfrac{e^{-i\frac{n\pi}{8}}(e^{-i\frac{n\pi}{8}}-e^{i\frac{n\pi}{8}})}{e^{i\frac{n\pi}{8}}(e^{i\frac{n\pi}{8}}-e^{-i\frac{n\pi}{8}})}\\ z_3&=\dfrac{-2i~\sin(\frac{n\pi}{8})~e^{-i\frac{n\pi}{8}}}{2i~\sin(\frac{n\pi}{8})~e^{i\frac{n\pi}{8}}}\\ z_3&=-e^{-i\left(\frac{n\pi}{4}\right)} \end{align*} On en dĂ©duit les formes gĂ©omĂ©triques et algĂ©briques: $$z_3=e^{i(\pi-\frac{n\pi}{4})}=-\cos(\frac{n\pi}{4})+i\sin(\frac{n\pi}{4})$$
  4. On a déjà vu que:
    $$(1+i)^n=2^{\frac{n}{2}}e^{i\frac{n\pi}{4}}$$ De mĂȘme on a: $$~~(1-i)^n=2^\frac{n}{2} e^{-i\frac{n\pi}{4}}$$ Par la suite: \begin{align*} (1+i)^n - (1-i)^n&=2^{\frac{n}{2}}(e^{i\frac{n\pi}{4}}-e^{-i\frac{n\pi}{4}})\\ &=2^{\left(\frac{n}{2}\right)}\left(2i~\sin\left(\frac{n\pi}{4}\right)\right)\\ &=i~2^{\left(\frac{n}{2}+1\right)}~\sin\left(\frac{n\pi}{4}\right) \end{align*} On en dĂ©duit: $$z_4=2^{\frac{n}{2}+1}\sin\left(\frac{n\pi}{4}\right)$$