Explanation
$$z_{1}$$ and $$z_{2}$$ are roots of $$z^{2}-az+a^{2}=0$$$$\therefore z_{1},z_{2}=\dfrac{a\pm \sqrt{a^{2}-4a^{2}}}{2}$$$$=a(\dfrac{1\pm i\sqrt{3}}{2})$$$$\therefore z_{1} , z_{2}$$ are multiple of complex cube roots of unity.$$\therefore |\dfrac{z_{1}}{z_{2}}|=|\dfrac{\omega ^{2}}{\omega }|=|\omega |=1$$
$$Z=\dfrac{(\dfrac{-1+\sqrt{3}i}{i})^{4n+1}}{(1-i\sqrt{3})^{4n}}$$
$$Z=\dfrac{(\dfrac{2\omega }{i})^{4n+1}}{(-2\omega )^{4n}}$$
$$=\dfrac{2\omega }{i^{4n+1}}\times (\dfrac{2\omega }{-2\omega })^{4n}$$
$$=\dfrac{2\omega }{i}\times 1$$
$$=\dfrac{-1+\sqrt{3}i}{i}$$
$$=\sqrt{3}+i$$$$\therefore arg |z|=tan^{-1}(\dfrac{1}{\sqrt{3}})$$$$=\dfrac{\pi }{6}$$
$$\Rightarrow\left | (x-a)+iy \right |=\left | (x+a)+iy \right |$$
$$\Rightarrow (x-a)^{2}+\not{ y^{2}} =(x+a)^{2}+\not{ y^{2}}$$
$$\Rightarrow\not{ x^{2}} +\not{ a^{2}} -2ax =\not{ x^{2}} +\not{ a^{2}} +2ax$$
$$\Rightarrow 0$$
$$\left |\dfrac{1}{(1-i)^{2}}-\dfrac{1}{(1+i)^{2}} \right |=\left |\dfrac{(1+i)^{2}-(1-i)^{2}}{\left ( (1-i)(1+i) \right )^{2}} \right |$$
$$=\left |\dfrac{\not{1}+2i-\not{1}-(\not{1}-2i)-\not{1} }{(1^{2}+1^{2})^{2}}\right |$$
$$=\left |\dfrac{4i}{4} \right |=1$$
$$a_{n}=(\sqrt{3+i})(\sqrt{3}-i)^{n-1}$$
$$\downarrow$$ $$\downarrow $$
first ratio
$$[as |z^{n-1}|=(|z|)^{n-1}]$$
$$\textbf{Step -1: Finding Discriminant of equation in option A.}$$
$$\text{The discriminant of any quadratic equation, }ax^2+bx+c=0, \text{ is }D=b^2-4ac.$$
$$\text{Discriminant of }x^2-2\sqrt3x+5=0\text{ is: }$$
$$D=(-2\sqrt3)^2-4\times1\times5$$
$$=12-20$$
$$=-8<0$$
$$\textbf{Step -2: Finding discriminant of equation in option B.}$$
$$\text{Discriminant of }2x^2+6\sqrt2x+9=0\text{ is: }$$
$$D=(6\sqrt2)^2-4\times2\times9$$
$$=72-72$$
$$=0$$
$$\textbf{Step -3: Finding discriminant of equation in option C.}$$
$$\text{Discriminant of }x^2-2\sqrt3x-5=0\text{ is: }$$
$$D=(-2\sqrt3)^2-4\times1\times-5$$
$$=12+20$$
$$=32>0$$
$$\textbf{Step -4: Finding discriminant of equation in option D.}$$
$$\text{Discriminant of }2x^2-6\sqrt2x-9=0\text{ is: }$$
$$D=(-6\sqrt2)^2-4\times2\times-9$$
$$=72+72$$
$$=144>0$$
$$\textbf{Step -5: Finding the correct option.}$$
$$\because \text{The equation in option A has discriminant less than zero.}$$
$$\Rightarrow \text{It has no real roots.}$$
$$\textbf{ Hence, option A is correct.}$$
$$\text{Discriminant of }-2x^2+6\sqrt2x+11=0\text{ is: }$$
$$D=(6\sqrt2)^2-4\times-2\times11$$
$$=72+88$$
$$=160>0$$
$$\text{Discriminant of }-2x^2-6\sqrt2x-9=0\text{ is: }$$
$$D=(-6\sqrt2)^2-4\times-2\times-9$$
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