A point z moves in the complex plane such that !(z-2z+2)=4; then the minimum value of |z-9…

Question

A point $z$ moves in the complex plane such that $\arg\!\left(\dfrac{z-2}{z+2}\right)=\dfrac\pi4$; then the minimum value of $|z-9\sqrt2-2i|^2$ equal to

Answer: 98

Accepted Answer

Similarly to Q206, this locus is the circle $x^2+y^2-4y-4=0$, i.e. centre $(0,2)$, radius $2\sqrt2$.

Distance from $(9\sqrt2,2)$ to the centre is $9\sqrt2$; the minimum distance to the circle is $9\sqrt2-2\sqrt2=7\sqrt2$, so the minimum of $|z-9\sqrt2-2i|^2$ is $(7\sqrt2)^2=98.$

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