What is the value of $$z^{2}+\frac{1}{z^{2}}$$ when $$z=5+2\sqrt{6}$$

Given :  $$z=5+2\sqrt{6}$$ ------------(i)

=> $$\frac{1}{z}=\frac{1}{5+2\sqrt{6}}$$

=> $$\frac{1}{z}=\frac{1}{5+2\sqrt{6}}\times(\frac{5-2\sqrt6}{5-2\sqrt6})$$

=> $$\frac{1}{z}=\frac{5-2\sqrt6}{25-24}=5-2\sqrt6$$ ------------(ii)

To find : $$z^{2}+\frac{1}{z^{2}}$$

Using, $$x^2+y^2=(x+y)^2-2xy$$

= $$(z+\frac{1}{z})^2-2(z)(\frac{1}{z})$$

Substituting values from equations (i) and (ii), we get :

= $$[(5+2\sqrt6)+(5-2\sqrt6)]^2-2(z)(\frac{1}{z})$$

= $$(10)^2-2=100-2=98$$

=> Ans - (C)