
\begin{enumerate}
\begin{multicols}{2}
\item $\begin{aligned}[t]
z_1 & = \dfrac{1}{2}\i z_0 + 1-\i \\
& = \dfrac{1}{2}\i \times \i + 1-\i \\
& = -\dfrac{1}{2} + 1-\i\\
& = \dfrac{1}{2}-\i.
\end{aligned}$

\columnbreak

$\begin{aligned}[t]
z_2 & = \dfrac{1}{2}\i z_1 + 1-\i \\
& =\dfrac{1}{2}\i \times \left( \dfrac{1}{2}-\i \right) + 1-\i \\
& = \dfrac{1}{4}\i + \dfrac{1}{2} + 1-\i\\
& = \dfrac{3}{2}-\dfrac{3}{4}\i.
\end{aligned}$
\end{multicols}

\begin{multicols}{2}
\item $\begin{aligned}[t]
v_{n+1} & = z_{n+1} -\dfrac{6}{5}+\dfrac{2}{5}\i\\
& = \dfrac{1}{2}\i z_n + 1-\i -\dfrac{6}{5}+\dfrac{2}{5}\i\\
& = \dfrac{1}{2}\i z_n -\dfrac{1}{5}-\dfrac{3}{5}\i\\
& = \dfrac{1}{2}\i\left(z_n-\dfrac{\tfrac{1}{5}}{\tfrac{1}{2}\i} - \dfrac{\tfrac{3}{5}\i}{\tfrac{1}{2}\i} \right)\\
& = \dfrac{1}{2}\i\left(z_n+\dfrac{2}{5}\i-\dfrac{6}{5}\right)\\
& = \dfrac{1}{2}\i v_n.
\end{aligned}$

\columnbreak

Ainsi, \suite{v}{n}{} est géométrique de raison\linebreak$q=\dfrac{1}{2}\i$ et de premier terme:
\begin{align*}
v_0 & = z_0-\dfrac{6}{5}+\dfrac{2}{5}\i\\
v_0 & = -\dfrac{6}{5}+\dfrac{7}{5}\i.
\end{align*}
\end{multicols}

\item On déduit de la question précédente que:
\[
v_n=v_0\times q^n = \left( -\dfrac{6}{5}+\dfrac{7}{5}\i \right) \times \left(\dfrac{1}{2}\i\right)^n
\]
et donc:
\[
z_n=v_n+\dfrac{6}{5}-\dfrac{2}{5}\i = \left( -\dfrac{6}{5}+\dfrac{7}{5}\i \right) \times \left(\dfrac{1}{2}\i\right)^n +\dfrac{6}{5}-\dfrac{2}{5}\i
\]

\item $|v_n|=|v_0| \times \left| \dfrac{1}{2}\i \right|^n = |v_0| \times \dfrac{1}{2^n} = \dfrac{|v_0|}{2^n}$.

Donc, $\limite{n}{+\infty}{|v_n|}=0$.

On peut alors en déduire que $\limite{n}{+\infty}{v_n}=0$.

Or, $z_n=v_n+\dfrac{6}{5}-\dfrac{2}{5}\i$ donc:
\[
\boxed{\limite{n}{+\infty}{z_n}=\dfrac{6}{5}-\dfrac{2}{5}\i}
\]
\end{enumerate}
