Plot $|fracz+iz-1|<1$

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Let $D=fracz+iz-1$ plot $D$



$$fracz+iz-1|<1iff frac<1iff |z+i|<|z-1|iff |x+(y+1)i|<|(x-1)+yi|iff\ iff sqrtx^2+(y+1)^2<sqrt(x-1)^2+y^2iff x^2+(y+1)^2<(x-1)^2+y^2iff \ iff x^2+y^2+2y+1<x^2-2x+1+y^2iff x<-y$$



Which is the are under the line $x=-y$



Is it correct?




complex-analysis




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    up vote
    2
    down vote

    favorite












    Let $D=fracz+iz-1$ plot $D$



    $$fracz+iz-1|<1iff frac<1iff |z+i|<|z-1|iff |x+(y+1)i|<|(x-1)+yi|iff\ iff sqrtx^2+(y+1)^2<sqrt(x-1)^2+y^2iff x^2+(y+1)^2<(x-1)^2+y^2iff \ iff x^2+y^2+2y+1<x^2-2x+1+y^2iff x<-y$$



    Which is the are under the line $x=-y$



    Is it correct?




    complex-analysis




    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $D=fracz+iz-1$ plot $D$



      $$fracz+iz-1|<1iff frac<1iff |z+i|<|z-1|iff |x+(y+1)i|<|(x-1)+yi|iff\ iff sqrtx^2+(y+1)^2<sqrt(x-1)^2+y^2iff x^2+(y+1)^2<(x-1)^2+y^2iff \ iff x^2+y^2+2y+1<x^2-2x+1+y^2iff x<-y$$



      Which is the are under the line $x=-y$



      Is it correct?




      complex-analysis




      share|cite|improve this question













      Let $D=fracz+iz-1$ plot $D$



      $$fracz+iz-1|<1iff frac<1iff |z+i|<|z-1|iff |x+(y+1)i|<|(x-1)+yi|iff\ iff sqrtx^2+(y+1)^2<sqrt(x-1)^2+y^2iff x^2+(y+1)^2<(x-1)^2+y^2iff \ iff x^2+y^2+2y+1<x^2-2x+1+y^2iff x<-y$$



      Which is the are under the line $x=-y$



      Is it correct?





      complex-analysis





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 15 at 8:04









      Siong Thye Goh

      77.8k134796




      77.8k134796









      asked Jul 15 at 7:56









      newhere

      759310




      759310




















          1 Answer
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          $$frac<1$$



          $$|z-(-i)| < |z-1|$$



          We are describing points that are closer to $(-i)$ than $1$.



          The perpendicular bisector between the two points is $y=-x$ and the corresponding region is the area under $y=-x$.






          share|cite|improve this answer





















          • So else from my mistake swapping $y<-x$ with $x<-y$ my way is correct?
            – newhere
            Jul 15 at 10:49










          • $y<-x$ and $x < -y$ are equivalent.
            – Siong Thye Goh
            Jul 15 at 12:27










          Your Answer




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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          4
          down vote



          accepted










          $$frac<1$$



          $$|z-(-i)| < |z-1|$$



          We are describing points that are closer to $(-i)$ than $1$.



          The perpendicular bisector between the two points is $y=-x$ and the corresponding region is the area under $y=-x$.






          share|cite|improve this answer





















          • So else from my mistake swapping $y<-x$ with $x<-y$ my way is correct?
            – newhere
            Jul 15 at 10:49










          • $y<-x$ and $x < -y$ are equivalent.
            – Siong Thye Goh
            Jul 15 at 12:27














          up vote
          4
          down vote



          accepted










          $$frac<1$$



          $$|z-(-i)| < |z-1|$$



          We are describing points that are closer to $(-i)$ than $1$.



          The perpendicular bisector between the two points is $y=-x$ and the corresponding region is the area under $y=-x$.






          share|cite|improve this answer





















          • So else from my mistake swapping $y<-x$ with $x<-y$ my way is correct?
            – newhere
            Jul 15 at 10:49










          • $y<-x$ and $x < -y$ are equivalent.
            – Siong Thye Goh
            Jul 15 at 12:27












          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          $$frac<1$$



          $$|z-(-i)| < |z-1|$$



          We are describing points that are closer to $(-i)$ than $1$.



          The perpendicular bisector between the two points is $y=-x$ and the corresponding region is the area under $y=-x$.






          share|cite|improve this answer













          $$frac<1$$



          $$|z-(-i)| < |z-1|$$



          We are describing points that are closer to $(-i)$ than $1$.



          The perpendicular bisector between the two points is $y=-x$ and the corresponding region is the area under $y=-x$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 15 at 8:03









          Siong Thye Goh

          77.8k134796




          77.8k134796











          • So else from my mistake swapping $y<-x$ with $x<-y$ my way is correct?
            – newhere
            Jul 15 at 10:49










          • $y<-x$ and $x < -y$ are equivalent.
            – Siong Thye Goh
            Jul 15 at 12:27
















          • So else from my mistake swapping $y<-x$ with $x<-y$ my way is correct?
            – newhere
            Jul 15 at 10:49










          • $y<-x$ and $x < -y$ are equivalent.
            – Siong Thye Goh
            Jul 15 at 12:27















          So else from my mistake swapping $y<-x$ with $x<-y$ my way is correct?
          – newhere
          Jul 15 at 10:49




          So else from my mistake swapping $y<-x$ with $x<-y$ my way is correct?
          – newhere
          Jul 15 at 10:49












          $y<-x$ and $x < -y$ are equivalent.
          – Siong Thye Goh
          Jul 15 at 12:27




          $y<-x$ and $x < -y$ are equivalent.
          – Siong Thye Goh
          Jul 15 at 12:27












           

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