If z is complex number such that $$|z|=\dfrac{1}{2}$$, then find the maximum value of $$|z^2-z+\dfrac{1}{4}|$$

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Uniteasy Admin · Accepted Answer

Let $$z=x+yi$$, then $$\sqrt{x^2+y^2}=\dfrac{1}{2}$$
$$x^2+y^2=\dfrac{1}{4} $$
z is a circle centered at origin.
$$|z^2-z+\dfrac{1}{4}|$$ 
$$=|z-\dfrac{1}{2}|^2 $$
$$=(x-\dfrac{1}{2})^2+y^2 $$
$$=x^2+y^2-x+\dfrac{1}{4} $$
$$=\dfrac{1}{2}-x $$
Since $$-\dfrac{1}{2} \leq x\leq \dfrac{1}{2} $$
The maximum value of $$|z^2-z+\dfrac{1}{4}|$$ happens when $$x=\dfrac{1}{2} $$, which is equal to $$1$$

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