Answer:
Step-by-step explanation:
a.
[tex]let~ z=\sqrt{1+\sqrt{3} i} \\also ~let~z=x+iy\\squaring\\z^2=x^2+i^2y^2+2 xy i\\\\or z^2=x^2-y^2+2xyi=1+\sqrt{3} i\\x^2-y^2=1\\2xy=2\sqrt{3} \\(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2\\(a+b)^2=(a-b)^2+4ab\\(x^2+y^2)^2=1^2+(2\sqrt{3} )^2\\(x^2+y^2)^2=1+12=13\\x^2+y^2=\sqrt{13} \\x^2-y^2=1\\2x^2=1+\sqrt{13} \\x=\pm\sqrt{\frac{1+\sqrt{13} }{2} } \\y=\frac{\sqrt{3} }{x} =\mp\sqrt{\frac{6}{1+\sqrt{13} } } \\so \sqrt{1+\sqrt{13} } =\pm\sqrt{\frac{1+\sqrt{13} }{2} } \mp\sqrt{\frac{6}{1+\sqrt{13} } } i[/tex]
