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.$$\therefore |\dfrac{z_{1}}{z_{2}}|=|\dfrac{\omega ^{2}}{\omega }|=|\omega |=1$$
$$\Rightarrow\left | (x-a)+iy \right |=\left | (x+a)+iy \right |$$
$$\Rightarrow (x-a)^{2}+\not{ y^{2}} =(x+a)^{2}+\not{ y^{2}}$$
$$\Rightarrow\not{ x^{2}} +\not{ a^{2}} -2ax =\not{ x^{2}} +\not{ a^{2}} +2ax$$
$$\Rightarrow 0$$
$$\left |\dfrac{1}{(1-i)^{2}}-\dfrac{1}{(1+i)^{2}} \right |=\left |\dfrac{(1+i)^{2}-(1-i)^{2}}{\left ( (1-i)(1+i) \right )^{2}} \right |$$
$$=\left |\dfrac{\not{1}+2i-\not{1}-(\not{1}-2i)-\not{1} }{(1^{2}+1^{2})^{2}}\right |$$
$$=\left |\dfrac{4i}{4} \right |=1$$
$$a_{n}=(\sqrt{3+i})(\sqrt{3}-i)^{n-1}$$
$$\downarrow$$ $$\downarrow $$
first ratio
$$[as |z^{n-1}|=(|z|)^{n-1}]$$
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