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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(2x−1)2+(2y)2=(x−1)2+y2x2−23x+19+y2=19(x−13)2+y2=(13)2⇒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.

∣2z−1∣=∣z−1∣⇒2∣z−12∣=∣z−1∣distanceto1isthedouble(notthesame!)distanceto0.5.

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 :)

thankyousirforpickingupmymistake:)

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