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Question-156400




Question Number 156400 by ajfour last updated on 10/Oct/21
Commented by ajfour last updated on 10/Oct/21
The roots of   f(z)=(z−a)^2 +b  is obtainable by the welded  parabola wires (with mutually  perpendicular planes;  upper  parallel Re-vertical, lower   one parallel to Im-Vertical).   Where they cross f(z)=c  gives  z=a±(√(c−b))   ★  (Today evening i imagined this!)    ...........................................
$$\mathrm{The}\:\mathrm{roots}\:\mathrm{of}\:\:\:\mathrm{f}\left(\mathrm{z}\right)=\left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{2}} +\mathrm{b} \\ $$$$\mathrm{is}\:\mathrm{obtainable}\:\mathrm{by}\:\mathrm{the}\:\mathrm{welded} \\ $$$$\mathrm{parabola}\:\mathrm{wires}\:\left(\mathrm{with}\:\mathrm{mutually}\right. \\ $$$$\mathrm{perpendicular}\:\mathrm{planes};\:\:\mathrm{upper} \\ $$$$\mathrm{parallel}\:\mathrm{Re}-\mathrm{vertical},\:\mathrm{lower}\: \\ $$$$\left.\mathrm{one}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{Im}-\mathrm{Vertical}\right).\: \\ $$$$\mathrm{Where}\:\mathrm{they}\:\mathrm{cross}\:\mathrm{f}\left(\mathrm{z}\right)=\mathrm{c} \\ $$$$\mathrm{gives}\:\:\mathrm{z}=\mathrm{a}\pm\sqrt{\mathrm{c}−\mathrm{b}}\:\:\:\bigstar \\ $$$$\left(\mathrm{Today}\:\mathrm{evening}\:\mathrm{i}\:\mathrm{imagined}\:\mathrm{this}!\right) \\ $$$$\:\:……………………………………. \\ $$

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