Hello,
[tex]Let's\ say \\\\z=\sqrt{3+4*i} =a+b*i\\\\z^2=3+4*i=(a+b*i)^2=a^2-b^2+2i*a*b\\\\\\if \ a\neq 0\\\left\{\begin{array}{ccc}a^2+b^2&=&3\\2ab=4\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}b=\dfrac{2}{a}\\a^2-(\dfrac{2}{a})^2=2\\\end{array}\right.\\\\\\a^4-4=3*a^2\\a^4-3a^2-4=0\\\\\Delta=(-3)^2-4*1*(-4)=25=5^2\\\\a^2=4\ or \ a^2=-1 (impossible)\\\\So:\\(a=2\ and\ b=1)\ or\ (a=-2\ and\ b=-1)\\[/tex]
Roots are thus 2+i and -2-i
There is an other using a geometrical formula (formule de Moivre)
