Question

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

Collected in the board: Imaginary number

Steven Zheng posted 2 days ago

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

Steven Zheng posted 2 days ago

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