A complex root

I believe there is much simpler solution. From the equations, we get that roots are z and -z. (x,-x) and (y,-y) Summation x^2 = 2x^2 Summation Y^2 = 2y^2.

Therefore, the question simplifies as 4 x^2 y^2 = 4/4 = 1.

To solve the square root of 2+ ${i}$ , firstly you write the square of your hypothetical answer and equate the real and imaginary parts, as follows: $(x+yi)^2$ = $x^{2}$ - $y^{2}$ +2 ${x}$ ${y}$ ${i}$ , so $x^{2}$ - $y^{2}$ = 2 ( 1) and 2 ${x}$ ${y}$ ${i}$ =1 (2 ). Rearranging (2 ) gives, ${y}$ = $\frac{1}{2x}$ which we substitute back into equation ( 1): $x^{2}$ - $\frac{1}{4x^2}$ = 2. Multiply by $4x^{2}$ and rearranging leads to 4 $x^{4}$ - 8 $x^{2}$ -1 = 0. This is a quadratic in disguise, so let ${t}$ = $x^{2}$ and solve 4 $t^{2}$ - 8 ${t}$ -1 = 0 (and square root the results) to produce the following 2 roots: $x_{1}$ = $\sqrt{1+\frac{1}{2}\sqrt{5}}$ and $x_{2}$ = $\sqrt{\frac{1}{2}\sqrt{5} -1}$ ${i}$ . Substitute the value of $x^{2}$ into equation (*1) yields: $y_{1}$ = $\sqrt{\frac{1}{2}\sqrt{5} -1}$ and $y_{2}$ = $\sqrt{1+\frac{1}{2}\sqrt{5}}$ ${i}$ . Now $\displaystyle \sum_{}^{}$ $x^{2}$ = 2 + $\sqrt{5}$ and $\displaystyle \sum_{}^{}$ $y^{2}$ = $\sqrt{5}$ -2. The answer is $\therefore$ ( $\sqrt{5}$ +2)( $\sqrt{5}$ -2) = 1