\begin{enumerate}
\item $|1+\i|=\sqrt{1^2+1^2}=\sqrt2$. Donc 
\[
1+\i=\sqrt2\left(\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt2}\i\right)=\sqrt2\left(\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}\i\right)=\sqrt2\left(\cos\dfrac{\pi}{4}+\i\sin\dfrac{\pi}{4}\right)=\boxed{\sqrt2\e^{\i\frac{\pi}{4}}}
\]
\item $\dfrac{1}{2}-\dfrac{\sqrt3}{2}\i=\cos\dfrac{\pi}{3}+\i\sin\dfrac{\pi}{3}=\boxed{\e^{\i\frac{\pi}{3}}}$

\item $|-1-\i\sqrt3|=\sqrt{(-1)^2+\big(-\sqrt3\big)^2}=\sqrt{4}=2$. Donc :
\[
-1-\i\sqrt3=2\left(-\dfrac{1}{2}-\dfrac{\sqrt3}{2}\i\right)=2\left[\cos\left(\pi+\dfrac{\pi}{3}\right)+\i\sin\left(\pi+\dfrac{\pi}{3}\right)\right]=\boxed{2\e^{\frac{4\pi}{3}\i}}
\]
\item $|2-2\i|=\sqrt{2^2+(-2)^2}=\sqrt8=2\sqrt2$. Donc :
\[
2-2\i=2\sqrt2\left(\dfrac{2}{2\sqrt2}-\dfrac{2}{2\sqrt2}\i\right)=2\sqrt2\left(\dfrac{\sqrt2}{2}-\dfrac{\sqrt2}{2}\i\right)=2\sqrt2\left[\cos\left(-\dfrac{\pi}{4}\right)+\i\sin\left(-\dfrac{\pi}{4}\right)\right]=\boxed{2\sqrt2\e^{-\frac{\pi}{4}\i}}
\]
\item $|2\sqrt3+2\i|=\sqrt{\big(2\sqrt3\big)^2+2^2}=\sqrt{12+4}=\sqrt{16}=4$. Donc :
\[
2\sqrt3+2\i=4\left(\dfrac{\sqrt3}{2}+\dfrac{1}{2}\i\right)=4\left(\cos\dfrac{\pi}{6}+\i\sin\dfrac{\pi}{6}\right)=\boxed{4\e^{\i\frac{\pi}{6}}}
\]
\item $\i=0+1\times\i=\cos\dfrac{\pi}{2}+\i\sin\dfrac{\pi}{2}=\boxed{\e^{\i\frac{\pi}{2}}}$
\end{enumerate}

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