How to plot ${z in mathbb{C} : |z-i|>|z+i|}$












2














How would I draw the set ${z in mathbb{C} : |z-i|>|z+i|}$ and ${z in mathbb{C} : |z-i|not=|z+i|}$?



Im not sure how to solve the second one, and for the first one, I tried squaring both sides and trying to work something out, but I got no where.



$|z-i|^2>|z+i|^2\\(z-i)(bar z+i)>(z+i)(bar z-i)\ zbar z+1+i(z -bar z)>z bar z+1 +i(bar z -z)\i(z-bar z)>i(bar z-z)$



What would the 'general' method/approach be for drawing the sets?



Edit: How would I draw ${z in mathbb{C} : |z-i|not=|z+i|}$?



After a similar calculation using Zev Chonoles' post, I got that $-bnot=b$, hence $z=a+ib$ satisfies $|z-i|not=|z+i|$ if and only if $-bnot=b$.



For ${z in mathbb{C} : |z-i|>|z+i|}$
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    2














    How would I draw the set ${z in mathbb{C} : |z-i|>|z+i|}$ and ${z in mathbb{C} : |z-i|not=|z+i|}$?



    Im not sure how to solve the second one, and for the first one, I tried squaring both sides and trying to work something out, but I got no where.



    $|z-i|^2>|z+i|^2\\(z-i)(bar z+i)>(z+i)(bar z-i)\ zbar z+1+i(z -bar z)>z bar z+1 +i(bar z -z)\i(z-bar z)>i(bar z-z)$



    What would the 'general' method/approach be for drawing the sets?



    Edit: How would I draw ${z in mathbb{C} : |z-i|not=|z+i|}$?



    After a similar calculation using Zev Chonoles' post, I got that $-bnot=b$, hence $z=a+ib$ satisfies $|z-i|not=|z+i|$ if and only if $-bnot=b$.



    For ${z in mathbb{C} : |z-i|>|z+i|}$
    1



    2






    complex-numbers







    share|cite|improve this question



























      2












      2








      2


      1





      How would I draw the set ${z in mathbb{C} : |z-i|>|z+i|}$ and ${z in mathbb{C} : |z-i|not=|z+i|}$?



      Im not sure how to solve the second one, and for the first one, I tried squaring both sides and trying to work something out, but I got no where.



      $|z-i|^2>|z+i|^2\\(z-i)(bar z+i)>(z+i)(bar z-i)\ zbar z+1+i(z -bar z)>z bar z+1 +i(bar z -z)\i(z-bar z)>i(bar z-z)$



      What would the 'general' method/approach be for drawing the sets?



      Edit: How would I draw ${z in mathbb{C} : |z-i|not=|z+i|}$?



      After a similar calculation using Zev Chonoles' post, I got that $-bnot=b$, hence $z=a+ib$ satisfies $|z-i|not=|z+i|$ if and only if $-bnot=b$.



      For ${z in mathbb{C} : |z-i|>|z+i|}$
      1



      2






      complex-numbers







      share|cite|improve this question















      How would I draw the set ${z in mathbb{C} : |z-i|>|z+i|}$ and ${z in mathbb{C} : |z-i|not=|z+i|}$?



      Im not sure how to solve the second one, and for the first one, I tried squaring both sides and trying to work something out, but I got no where.



      $|z-i|^2>|z+i|^2\\(z-i)(bar z+i)>(z+i)(bar z-i)\ zbar z+1+i(z -bar z)>z bar z+1 +i(bar z -z)\i(z-bar z)>i(bar z-z)$



      What would the 'general' method/approach be for drawing the sets?



      Edit: How would I draw ${z in mathbb{C} : |z-i|not=|z+i|}$?



      After a similar calculation using Zev Chonoles' post, I got that $-bnot=b$, hence $z=a+ib$ satisfies $|z-i|not=|z+i|$ if and only if $-bnot=b$.



      For ${z in mathbb{C} : |z-i|>|z+i|}$
      1



      2






      complex-numbers




      complex-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 19:44









      Glorfindel

      3,41981830




      3,41981830










      asked May 23 '12 at 22:45









      DerrickDerrick

      3461714




      3461714






















          3 Answers
          3






          active

          oldest

          votes


















          5














          Write out a complex number $z$ with real and imaginary components, i.e. as $z=a+bi$. Then
          $$|z-i|=|a+(b-1)i|=sqrt{a^2+(b-1)^2}$$
          $$|z+i|=|a+(b+1)i|=sqrt{a^2+(b+1)^2}$$
          so
          $$begin{align*}|z-i|>|z+i|&iffsqrt{a^2+(b-1)^2}>sqrt{a^2+(b+1)^2}\&iff a^2+(b-1)^2>a^2+(b+1)^2\ &iff (b-1)^2>(b+1)^2\ &iff -2b>2b\ &iff b<0end{align*}$$
          Thus, the complex number $z=a+bi$ satisfies $|z-i|>|z+i|$ if and only if $b<0$, i.e. if and only if it lies below the real axis in the complex plane. Thinking geometrically (i.e. with complex numbers as points in the plane), it might also help to note that
          $$|z-i|=sqrt{a^2+(b-1)^2}=text{distance from }(a,b)text{ to }(0,1)$$
          $$|z+i|=sqrt{a^2+(b+1)^2}=text{distance from }(a,b)text{ to }(0,-1)$$






          share|cite|improve this answer





















          • Thanks, @ZevChonoles . Just to make sure, is my diagram correct for the first graph? Also, how would I draw the 2nd graph?
            – Derrick
            May 23 '12 at 23:11






          • 1




            @Derrick: Yup, it's correct - you have the dotted line along the real axis indicating that it is not included in the set, and you've shaded in the right region. Now, note that $-bneq b$ if and only if $bneq 0$. Does that help you see the graph for the second question?
            – Zev Chonoles
            May 23 '12 at 23:12












          • Thanks @ZevChonoles, so I take it, it would be shading the entire region with a dotted line along the real axis (see edited post with new picture)?
            – Derrick
            May 23 '12 at 23:19






          • 1




            @Derrick: Exactly!
            – Zev Chonoles
            May 23 '12 at 23:32










          • Okay, thanks again for all your help!
            – Derrick
            May 23 '12 at 23:33



















          2














          Try writing $z=x+iy$ with $x,yinmathbb{R}$.



          Then for the first inequality you get (just try it for yourself):



          $$|z-i|>|z+i| Leftrightarrow mbox{Im}(z) < 0$$



          so the solutions is the whole lower half-plane (without the real axis).



          For the second one, you get $mathbb{C}setminusmathbb{R}$, because of a similar condition for the imaginary part ($mbox{Im}(z)not=0$).



          Also, your last line $i(z-bar{z})>i(bar{z}-z)$ translates for $z=x+iy$ into $0>y$, which gives the same result.






          share|cite|improve this answer





















          • Thanks, most appreciated :)
            – Derrick
            May 23 '12 at 23:34



















          1














          One could simply look at the absolute value of a complex number as it's distance from origin, or in other words, $|a-b|$ is the distance of $a$ from $b$.



          Now, in your problem you want find all the points such that their distance from $i$ is more than their distance from $-i$. Well, find all the points who have the same distance from $i$ and $-i$. That's a line (namely the line that is perpendicular to the line segment $overline{i,-i}$).



          So far you've got the solution to the second part of the problem. Now, choose the half-plane which includes $-i$. That is that answer for the first part.






          share|cite|improve this answer





















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            3 Answers
            3






            active

            oldest

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            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5














            Write out a complex number $z$ with real and imaginary components, i.e. as $z=a+bi$. Then
            $$|z-i|=|a+(b-1)i|=sqrt{a^2+(b-1)^2}$$
            $$|z+i|=|a+(b+1)i|=sqrt{a^2+(b+1)^2}$$
            so
            $$begin{align*}|z-i|>|z+i|&iffsqrt{a^2+(b-1)^2}>sqrt{a^2+(b+1)^2}\&iff a^2+(b-1)^2>a^2+(b+1)^2\ &iff (b-1)^2>(b+1)^2\ &iff -2b>2b\ &iff b<0end{align*}$$
            Thus, the complex number $z=a+bi$ satisfies $|z-i|>|z+i|$ if and only if $b<0$, i.e. if and only if it lies below the real axis in the complex plane. Thinking geometrically (i.e. with complex numbers as points in the plane), it might also help to note that
            $$|z-i|=sqrt{a^2+(b-1)^2}=text{distance from }(a,b)text{ to }(0,1)$$
            $$|z+i|=sqrt{a^2+(b+1)^2}=text{distance from }(a,b)text{ to }(0,-1)$$






            share|cite|improve this answer





















            • Thanks, @ZevChonoles . Just to make sure, is my diagram correct for the first graph? Also, how would I draw the 2nd graph?
              – Derrick
              May 23 '12 at 23:11






            • 1




              @Derrick: Yup, it's correct - you have the dotted line along the real axis indicating that it is not included in the set, and you've shaded in the right region. Now, note that $-bneq b$ if and only if $bneq 0$. Does that help you see the graph for the second question?
              – Zev Chonoles
              May 23 '12 at 23:12












            • Thanks @ZevChonoles, so I take it, it would be shading the entire region with a dotted line along the real axis (see edited post with new picture)?
              – Derrick
              May 23 '12 at 23:19






            • 1




              @Derrick: Exactly!
              – Zev Chonoles
              May 23 '12 at 23:32










            • Okay, thanks again for all your help!
              – Derrick
              May 23 '12 at 23:33
















            5














            Write out a complex number $z$ with real and imaginary components, i.e. as $z=a+bi$. Then
            $$|z-i|=|a+(b-1)i|=sqrt{a^2+(b-1)^2}$$
            $$|z+i|=|a+(b+1)i|=sqrt{a^2+(b+1)^2}$$
            so
            $$begin{align*}|z-i|>|z+i|&iffsqrt{a^2+(b-1)^2}>sqrt{a^2+(b+1)^2}\&iff a^2+(b-1)^2>a^2+(b+1)^2\ &iff (b-1)^2>(b+1)^2\ &iff -2b>2b\ &iff b<0end{align*}$$
            Thus, the complex number $z=a+bi$ satisfies $|z-i|>|z+i|$ if and only if $b<0$, i.e. if and only if it lies below the real axis in the complex plane. Thinking geometrically (i.e. with complex numbers as points in the plane), it might also help to note that
            $$|z-i|=sqrt{a^2+(b-1)^2}=text{distance from }(a,b)text{ to }(0,1)$$
            $$|z+i|=sqrt{a^2+(b+1)^2}=text{distance from }(a,b)text{ to }(0,-1)$$






            share|cite|improve this answer





















            • Thanks, @ZevChonoles . Just to make sure, is my diagram correct for the first graph? Also, how would I draw the 2nd graph?
              – Derrick
              May 23 '12 at 23:11






            • 1




              @Derrick: Yup, it's correct - you have the dotted line along the real axis indicating that it is not included in the set, and you've shaded in the right region. Now, note that $-bneq b$ if and only if $bneq 0$. Does that help you see the graph for the second question?
              – Zev Chonoles
              May 23 '12 at 23:12












            • Thanks @ZevChonoles, so I take it, it would be shading the entire region with a dotted line along the real axis (see edited post with new picture)?
              – Derrick
              May 23 '12 at 23:19






            • 1




              @Derrick: Exactly!
              – Zev Chonoles
              May 23 '12 at 23:32










            • Okay, thanks again for all your help!
              – Derrick
              May 23 '12 at 23:33














            5












            5








            5






            Write out a complex number $z$ with real and imaginary components, i.e. as $z=a+bi$. Then
            $$|z-i|=|a+(b-1)i|=sqrt{a^2+(b-1)^2}$$
            $$|z+i|=|a+(b+1)i|=sqrt{a^2+(b+1)^2}$$
            so
            $$begin{align*}|z-i|>|z+i|&iffsqrt{a^2+(b-1)^2}>sqrt{a^2+(b+1)^2}\&iff a^2+(b-1)^2>a^2+(b+1)^2\ &iff (b-1)^2>(b+1)^2\ &iff -2b>2b\ &iff b<0end{align*}$$
            Thus, the complex number $z=a+bi$ satisfies $|z-i|>|z+i|$ if and only if $b<0$, i.e. if and only if it lies below the real axis in the complex plane. Thinking geometrically (i.e. with complex numbers as points in the plane), it might also help to note that
            $$|z-i|=sqrt{a^2+(b-1)^2}=text{distance from }(a,b)text{ to }(0,1)$$
            $$|z+i|=sqrt{a^2+(b+1)^2}=text{distance from }(a,b)text{ to }(0,-1)$$






            share|cite|improve this answer












            Write out a complex number $z$ with real and imaginary components, i.e. as $z=a+bi$. Then
            $$|z-i|=|a+(b-1)i|=sqrt{a^2+(b-1)^2}$$
            $$|z+i|=|a+(b+1)i|=sqrt{a^2+(b+1)^2}$$
            so
            $$begin{align*}|z-i|>|z+i|&iffsqrt{a^2+(b-1)^2}>sqrt{a^2+(b+1)^2}\&iff a^2+(b-1)^2>a^2+(b+1)^2\ &iff (b-1)^2>(b+1)^2\ &iff -2b>2b\ &iff b<0end{align*}$$
            Thus, the complex number $z=a+bi$ satisfies $|z-i|>|z+i|$ if and only if $b<0$, i.e. if and only if it lies below the real axis in the complex plane. Thinking geometrically (i.e. with complex numbers as points in the plane), it might also help to note that
            $$|z-i|=sqrt{a^2+(b-1)^2}=text{distance from }(a,b)text{ to }(0,1)$$
            $$|z+i|=sqrt{a^2+(b+1)^2}=text{distance from }(a,b)text{ to }(0,-1)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered May 23 '12 at 22:52









            Zev ChonolesZev Chonoles

            109k16226423




            109k16226423












            • Thanks, @ZevChonoles . Just to make sure, is my diagram correct for the first graph? Also, how would I draw the 2nd graph?
              – Derrick
              May 23 '12 at 23:11






            • 1




              @Derrick: Yup, it's correct - you have the dotted line along the real axis indicating that it is not included in the set, and you've shaded in the right region. Now, note that $-bneq b$ if and only if $bneq 0$. Does that help you see the graph for the second question?
              – Zev Chonoles
              May 23 '12 at 23:12












            • Thanks @ZevChonoles, so I take it, it would be shading the entire region with a dotted line along the real axis (see edited post with new picture)?
              – Derrick
              May 23 '12 at 23:19






            • 1




              @Derrick: Exactly!
              – Zev Chonoles
              May 23 '12 at 23:32










            • Okay, thanks again for all your help!
              – Derrick
              May 23 '12 at 23:33


















            • Thanks, @ZevChonoles . Just to make sure, is my diagram correct for the first graph? Also, how would I draw the 2nd graph?
              – Derrick
              May 23 '12 at 23:11






            • 1




              @Derrick: Yup, it's correct - you have the dotted line along the real axis indicating that it is not included in the set, and you've shaded in the right region. Now, note that $-bneq b$ if and only if $bneq 0$. Does that help you see the graph for the second question?
              – Zev Chonoles
              May 23 '12 at 23:12












            • Thanks @ZevChonoles, so I take it, it would be shading the entire region with a dotted line along the real axis (see edited post with new picture)?
              – Derrick
              May 23 '12 at 23:19






            • 1




              @Derrick: Exactly!
              – Zev Chonoles
              May 23 '12 at 23:32










            • Okay, thanks again for all your help!
              – Derrick
              May 23 '12 at 23:33
















            Thanks, @ZevChonoles . Just to make sure, is my diagram correct for the first graph? Also, how would I draw the 2nd graph?
            – Derrick
            May 23 '12 at 23:11




            Thanks, @ZevChonoles . Just to make sure, is my diagram correct for the first graph? Also, how would I draw the 2nd graph?
            – Derrick
            May 23 '12 at 23:11




            1




            1




            @Derrick: Yup, it's correct - you have the dotted line along the real axis indicating that it is not included in the set, and you've shaded in the right region. Now, note that $-bneq b$ if and only if $bneq 0$. Does that help you see the graph for the second question?
            – Zev Chonoles
            May 23 '12 at 23:12






            @Derrick: Yup, it's correct - you have the dotted line along the real axis indicating that it is not included in the set, and you've shaded in the right region. Now, note that $-bneq b$ if and only if $bneq 0$. Does that help you see the graph for the second question?
            – Zev Chonoles
            May 23 '12 at 23:12














            Thanks @ZevChonoles, so I take it, it would be shading the entire region with a dotted line along the real axis (see edited post with new picture)?
            – Derrick
            May 23 '12 at 23:19




            Thanks @ZevChonoles, so I take it, it would be shading the entire region with a dotted line along the real axis (see edited post with new picture)?
            – Derrick
            May 23 '12 at 23:19




            1




            1




            @Derrick: Exactly!
            – Zev Chonoles
            May 23 '12 at 23:32




            @Derrick: Exactly!
            – Zev Chonoles
            May 23 '12 at 23:32












            Okay, thanks again for all your help!
            – Derrick
            May 23 '12 at 23:33




            Okay, thanks again for all your help!
            – Derrick
            May 23 '12 at 23:33











            2














            Try writing $z=x+iy$ with $x,yinmathbb{R}$.



            Then for the first inequality you get (just try it for yourself):



            $$|z-i|>|z+i| Leftrightarrow mbox{Im}(z) < 0$$



            so the solutions is the whole lower half-plane (without the real axis).



            For the second one, you get $mathbb{C}setminusmathbb{R}$, because of a similar condition for the imaginary part ($mbox{Im}(z)not=0$).



            Also, your last line $i(z-bar{z})>i(bar{z}-z)$ translates for $z=x+iy$ into $0>y$, which gives the same result.






            share|cite|improve this answer





















            • Thanks, most appreciated :)
              – Derrick
              May 23 '12 at 23:34
















            2














            Try writing $z=x+iy$ with $x,yinmathbb{R}$.



            Then for the first inequality you get (just try it for yourself):



            $$|z-i|>|z+i| Leftrightarrow mbox{Im}(z) < 0$$



            so the solutions is the whole lower half-plane (without the real axis).



            For the second one, you get $mathbb{C}setminusmathbb{R}$, because of a similar condition for the imaginary part ($mbox{Im}(z)not=0$).



            Also, your last line $i(z-bar{z})>i(bar{z}-z)$ translates for $z=x+iy$ into $0>y$, which gives the same result.






            share|cite|improve this answer





















            • Thanks, most appreciated :)
              – Derrick
              May 23 '12 at 23:34














            2












            2








            2






            Try writing $z=x+iy$ with $x,yinmathbb{R}$.



            Then for the first inequality you get (just try it for yourself):



            $$|z-i|>|z+i| Leftrightarrow mbox{Im}(z) < 0$$



            so the solutions is the whole lower half-plane (without the real axis).



            For the second one, you get $mathbb{C}setminusmathbb{R}$, because of a similar condition for the imaginary part ($mbox{Im}(z)not=0$).



            Also, your last line $i(z-bar{z})>i(bar{z}-z)$ translates for $z=x+iy$ into $0>y$, which gives the same result.






            share|cite|improve this answer












            Try writing $z=x+iy$ with $x,yinmathbb{R}$.



            Then for the first inequality you get (just try it for yourself):



            $$|z-i|>|z+i| Leftrightarrow mbox{Im}(z) < 0$$



            so the solutions is the whole lower half-plane (without the real axis).



            For the second one, you get $mathbb{C}setminusmathbb{R}$, because of a similar condition for the imaginary part ($mbox{Im}(z)not=0$).



            Also, your last line $i(z-bar{z})>i(bar{z}-z)$ translates for $z=x+iy$ into $0>y$, which gives the same result.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered May 23 '12 at 22:56









            Gregor BoteroGregor Botero

            2,29411530




            2,29411530












            • Thanks, most appreciated :)
              – Derrick
              May 23 '12 at 23:34


















            • Thanks, most appreciated :)
              – Derrick
              May 23 '12 at 23:34
















            Thanks, most appreciated :)
            – Derrick
            May 23 '12 at 23:34




            Thanks, most appreciated :)
            – Derrick
            May 23 '12 at 23:34











            1














            One could simply look at the absolute value of a complex number as it's distance from origin, or in other words, $|a-b|$ is the distance of $a$ from $b$.



            Now, in your problem you want find all the points such that their distance from $i$ is more than their distance from $-i$. Well, find all the points who have the same distance from $i$ and $-i$. That's a line (namely the line that is perpendicular to the line segment $overline{i,-i}$).



            So far you've got the solution to the second part of the problem. Now, choose the half-plane which includes $-i$. That is that answer for the first part.






            share|cite|improve this answer


























              1














              One could simply look at the absolute value of a complex number as it's distance from origin, or in other words, $|a-b|$ is the distance of $a$ from $b$.



              Now, in your problem you want find all the points such that their distance from $i$ is more than their distance from $-i$. Well, find all the points who have the same distance from $i$ and $-i$. That's a line (namely the line that is perpendicular to the line segment $overline{i,-i}$).



              So far you've got the solution to the second part of the problem. Now, choose the half-plane which includes $-i$. That is that answer for the first part.






              share|cite|improve this answer
























                1












                1








                1






                One could simply look at the absolute value of a complex number as it's distance from origin, or in other words, $|a-b|$ is the distance of $a$ from $b$.



                Now, in your problem you want find all the points such that their distance from $i$ is more than their distance from $-i$. Well, find all the points who have the same distance from $i$ and $-i$. That's a line (namely the line that is perpendicular to the line segment $overline{i,-i}$).



                So far you've got the solution to the second part of the problem. Now, choose the half-plane which includes $-i$. That is that answer for the first part.






                share|cite|improve this answer












                One could simply look at the absolute value of a complex number as it's distance from origin, or in other words, $|a-b|$ is the distance of $a$ from $b$.



                Now, in your problem you want find all the points such that their distance from $i$ is more than their distance from $-i$. Well, find all the points who have the same distance from $i$ and $-i$. That's a line (namely the line that is perpendicular to the line segment $overline{i,-i}$).



                So far you've got the solution to the second part of the problem. Now, choose the half-plane which includes $-i$. That is that answer for the first part.







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                answered May 23 '12 at 23:46









                KeivanKeivan

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