On sait que, pour tout nombre complexe \(\displaystyle{z= a+ib}\) :
\(\displaystyle{\left| z \right| = \sqrt{a^2+b^2}}\)
Ici, on obtient :
\(\displaystyle{\left| z \right| = \sqrt{\left(\sqrt3\right)^2+3^2}}\)
\(\displaystyle{\left| z \right| = \sqrt{ 3+9}}\)
\(\displaystyle{\left| z \right| = \sqrt{12}}\)
Finalement :
\(\displaystyle{\left| z \right| = 2\sqrt{ 3}}\)