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Question Number 28219 by math solver last updated on 22/Jan/18

Commented by mrW2 last updated on 22/Jan/18

z=x+iy  (2x−1)^2 +(2y)^2 =(x−1)^2 +y^2   x^2 −(2/3)x+(1/9)+y^2 =(1/9)  (x−(1/3))^2 +y^2 =((1/3))^2   ⇒circle

$${z}={x}+{iy} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{y}\right)^{\mathrm{2}} =\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{x}+\frac{\mathrm{1}}{\mathrm{9}}+{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{circle} \\ $$

Commented by mrW2 last updated on 22/Jan/18

∣2z−1∣=∣z−1∣  ⇒2∣z−(1/2)∣=∣z−1∣  distance to 1 is the double (not the same!) distance to 0.5.

$$\mid\mathrm{2}{z}−\mathrm{1}\mid=\mid{z}−\mathrm{1}\mid \\ $$$$\Rightarrow\mathrm{2}\mid{z}−\frac{\mathrm{1}}{\mathrm{2}}\mid=\mid{z}−\mathrm{1}\mid \\ $$$${distance}\:{to}\:\mathrm{1}\:{is}\:{the}\:{double}\:\left({not}\:{the}\:{same}!\right)\:{distance}\:{to}\:\mathrm{0}.\mathrm{5}. \\ $$

Commented by ajfour last updated on 22/Jan/18

circle

$${circle} \\ $$

Commented by math solver last updated on 22/Jan/18

thank you sir for picking up  my mistake :)

$${thank}\:{you}\:{sir}\:{for}\:{picking}\:{up} \\ $$$$\left.{my}\:{mistake}\::\right) \\ $$

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