Solving complex equation: $(z-1)^2+(bar{z}-2i)^2 = 0$











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We're supposed to solve this complex numbers equation:



$(z-1)^2+(bar{z}-2i)^2 = 0$



I'm getting the result:



$z_{1} = frac{1-i}{2}, z_{2} = frac{1+i}{2}$



Others are getting the same result. However, the answers page says that the result should be:



$z_{0} = - frac{3}{10} + frac{3}{5}i$



Would anyone be able to give it a look and verify whether I am wrong or the answers page is wrong? Thanks.










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  • 4




    Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
    – amWhy
    Nov 19 at 18:27








  • 1




    You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
    – Adrian Keister
    Nov 19 at 18:54















up vote
0
down vote

favorite
1












We're supposed to solve this complex numbers equation:



$(z-1)^2+(bar{z}-2i)^2 = 0$



I'm getting the result:



$z_{1} = frac{1-i}{2}, z_{2} = frac{1+i}{2}$



Others are getting the same result. However, the answers page says that the result should be:



$z_{0} = - frac{3}{10} + frac{3}{5}i$



Would anyone be able to give it a look and verify whether I am wrong or the answers page is wrong? Thanks.










share|cite|improve this question




















  • 4




    Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
    – amWhy
    Nov 19 at 18:27








  • 1




    You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
    – Adrian Keister
    Nov 19 at 18:54













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





We're supposed to solve this complex numbers equation:



$(z-1)^2+(bar{z}-2i)^2 = 0$



I'm getting the result:



$z_{1} = frac{1-i}{2}, z_{2} = frac{1+i}{2}$



Others are getting the same result. However, the answers page says that the result should be:



$z_{0} = - frac{3}{10} + frac{3}{5}i$



Would anyone be able to give it a look and verify whether I am wrong or the answers page is wrong? Thanks.










share|cite|improve this question















We're supposed to solve this complex numbers equation:



$(z-1)^2+(bar{z}-2i)^2 = 0$



I'm getting the result:



$z_{1} = frac{1-i}{2}, z_{2} = frac{1+i}{2}$



Others are getting the same result. However, the answers page says that the result should be:



$z_{0} = - frac{3}{10} + frac{3}{5}i$



Would anyone be able to give it a look and verify whether I am wrong or the answers page is wrong? Thanks.







complex-numbers






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edited Nov 19 at 18:39









amWhy

191k27223439




191k27223439










asked Nov 19 at 18:25









weno

415




415








  • 4




    Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
    – amWhy
    Nov 19 at 18:27








  • 1




    You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
    – Adrian Keister
    Nov 19 at 18:54














  • 4




    Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
    – amWhy
    Nov 19 at 18:27








  • 1




    You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
    – Adrian Keister
    Nov 19 at 18:54








4




4




Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
– amWhy
Nov 19 at 18:27






Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
– amWhy
Nov 19 at 18:27






1




1




You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
– Adrian Keister
Nov 19 at 18:54




You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
– Adrian Keister
Nov 19 at 18:54










4 Answers
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There is no solution for the complex equation below is the proof:
$(z-1)^2+(bar{z}-2i)^2 = 0 Rightarrow z^2+1-2z+bar{z^2}-4-4ibar{z}=0$
We know $z^2+bar{z^2}=2 Re{z^2}=2x^2-2y^2$ where $z=x+iy$ and $x$ and $y$ are real numbers. Substituting $z=x+iy$ in the complex equation we get $2x^2-2y^2-3-2(x+iy)-4i(x-iy)=0$. So both real and imaginary parts of the resultant equation must be zero. So we get:
$$2x^2-2y^2-2x-4y-3=0$$
and
$$-2y-4x=0 Rightarrow y=-2x$$
Substituting $y=-2x$ into the first real equation we get
$$2x^2-8x^2-2x+8x-3=0$$
or$$-6x^2+6x-3=0$$
or $$x^2-x+0.5=0$$
so $$x=frac{1pm i}{2}$$ which is not a real number. So there will not be any real $x$ and $y$ satisfying the complex equation. Therefore no complex $z$ satisfies the equation.






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    up vote
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    $z=frac{-3+6i}{10}$ is not a solution to this problem. Let's make sure.



    $z-1=frac{-13+6i}{10}$,



    $bar{z}-2i=frac{-3-6i}{10}-frac{20i}{10} =frac{-3-26i}{10}$



    $$(z-1)^2=frac{(-13+6i)^2}{100}=frac{169 - 156i-36}{100}=frac{133-156i}{100}$$



    $$(bar{z}-2i)^2 =frac{(-3-26i)^2}{100}=frac{9-156i-676}{100}=frac{156i-667}{100}$$



    So therefore $(z-1)^2+(bar{z}-2i)^2=-5.34$



    In fact, there are not solutions to this equation.



    It's not that hard to see if you apply $a^2+b^2=(a+bi)(a-bi)$
    Taking $a=z-1$ and $b=bar{z}-2i implies bi= bar{z}i+2$



    $a+bi=z-1+bar{z}+2=z+bar{z}+1$ and



    $a-bi=z-1-bar{z}-2=z-bar{z}-3$ and



    We arrive at



    $(z-1)^2+(bar{z}-2i)^2=(z+bar{z}i+1)(z-bar{z}i-3)=0$



    Taking $z=a+bi implies bar{z}=a-biimplies bar{z}i=b+ai$



    So then $z+bar{z}i=(a+b)+(a+b)i$ and $z-bar{z}i= a-b-(a-b)i$
    What's note worthy here is that the real and the imaginary parts of these numbers are the same.



    Then $z+bar{z}i=-1 implies a+b+(a+b)i=-1$ but then $a+b$ is simulataneously $0$ and $-1$ and likewise $z-bar{z}i=3implies a-b-(a-b)i=3$ implies that the $(a-b)$ is simultaneously $3$ and $0$.



    So we conclude: No solutions.






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      up vote
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      down vote













      Alternative way to arrive to the conclusion that the equation has no solutions:



      Write $;z=a+biimpliesoverline z=a-bi;$ , and the given equation is



      $$(a-1+bi)^2+(a-(b+2)i)^2=0iff$$



      $$(a-1)^2-b^2+2(a-1)bi+a^2-(b+2)^2-2a(b+2)i=0$$



      and compare now real and imaginary parts:



      $$begin{cases}(a-1)^2-b^2+a^2-(b+2)^2=0iff2a^2-2b^2-2a-4b-3=0\
      2ab-2b-2ab-4a=0iff b=-2aend{cases};;;implies$$



      $$2a^2-8a^2-2a+8a-3=0iff 6a^2-6a+3=0$$



      adn the last equation has no real solution...as it should if there was a solution.






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        Late answer but I think worth mentioning it:



        You can write the expression as difference of squares and factor it into two factors which can never be $0$:
        $$begin{eqnarray*}(z-1)^2+(bar{z}-2i)^2 & = & (z-1)^2-i^2(bar{z}-2i)^2 \
        & = & (z+1 + ibar z)(z-3 - ibar z) \
        & = & 0
        end{eqnarray*}$$

        With $z = a+ib$ you get




        • $z+1 + ibar z = a+b+1 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$

        • $z-3 - ibar z =a+b-3 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$






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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

          oldest

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          active

          oldest

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          active

          oldest

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          up vote
          0
          down vote



          accepted










          There is no solution for the complex equation below is the proof:
          $(z-1)^2+(bar{z}-2i)^2 = 0 Rightarrow z^2+1-2z+bar{z^2}-4-4ibar{z}=0$
          We know $z^2+bar{z^2}=2 Re{z^2}=2x^2-2y^2$ where $z=x+iy$ and $x$ and $y$ are real numbers. Substituting $z=x+iy$ in the complex equation we get $2x^2-2y^2-3-2(x+iy)-4i(x-iy)=0$. So both real and imaginary parts of the resultant equation must be zero. So we get:
          $$2x^2-2y^2-2x-4y-3=0$$
          and
          $$-2y-4x=0 Rightarrow y=-2x$$
          Substituting $y=-2x$ into the first real equation we get
          $$2x^2-8x^2-2x+8x-3=0$$
          or$$-6x^2+6x-3=0$$
          or $$x^2-x+0.5=0$$
          so $$x=frac{1pm i}{2}$$ which is not a real number. So there will not be any real $x$ and $y$ satisfying the complex equation. Therefore no complex $z$ satisfies the equation.






          share|cite|improve this answer

























            up vote
            0
            down vote



            accepted










            There is no solution for the complex equation below is the proof:
            $(z-1)^2+(bar{z}-2i)^2 = 0 Rightarrow z^2+1-2z+bar{z^2}-4-4ibar{z}=0$
            We know $z^2+bar{z^2}=2 Re{z^2}=2x^2-2y^2$ where $z=x+iy$ and $x$ and $y$ are real numbers. Substituting $z=x+iy$ in the complex equation we get $2x^2-2y^2-3-2(x+iy)-4i(x-iy)=0$. So both real and imaginary parts of the resultant equation must be zero. So we get:
            $$2x^2-2y^2-2x-4y-3=0$$
            and
            $$-2y-4x=0 Rightarrow y=-2x$$
            Substituting $y=-2x$ into the first real equation we get
            $$2x^2-8x^2-2x+8x-3=0$$
            or$$-6x^2+6x-3=0$$
            or $$x^2-x+0.5=0$$
            so $$x=frac{1pm i}{2}$$ which is not a real number. So there will not be any real $x$ and $y$ satisfying the complex equation. Therefore no complex $z$ satisfies the equation.






            share|cite|improve this answer























              up vote
              0
              down vote



              accepted







              up vote
              0
              down vote



              accepted






              There is no solution for the complex equation below is the proof:
              $(z-1)^2+(bar{z}-2i)^2 = 0 Rightarrow z^2+1-2z+bar{z^2}-4-4ibar{z}=0$
              We know $z^2+bar{z^2}=2 Re{z^2}=2x^2-2y^2$ where $z=x+iy$ and $x$ and $y$ are real numbers. Substituting $z=x+iy$ in the complex equation we get $2x^2-2y^2-3-2(x+iy)-4i(x-iy)=0$. So both real and imaginary parts of the resultant equation must be zero. So we get:
              $$2x^2-2y^2-2x-4y-3=0$$
              and
              $$-2y-4x=0 Rightarrow y=-2x$$
              Substituting $y=-2x$ into the first real equation we get
              $$2x^2-8x^2-2x+8x-3=0$$
              or$$-6x^2+6x-3=0$$
              or $$x^2-x+0.5=0$$
              so $$x=frac{1pm i}{2}$$ which is not a real number. So there will not be any real $x$ and $y$ satisfying the complex equation. Therefore no complex $z$ satisfies the equation.






              share|cite|improve this answer












              There is no solution for the complex equation below is the proof:
              $(z-1)^2+(bar{z}-2i)^2 = 0 Rightarrow z^2+1-2z+bar{z^2}-4-4ibar{z}=0$
              We know $z^2+bar{z^2}=2 Re{z^2}=2x^2-2y^2$ where $z=x+iy$ and $x$ and $y$ are real numbers. Substituting $z=x+iy$ in the complex equation we get $2x^2-2y^2-3-2(x+iy)-4i(x-iy)=0$. So both real and imaginary parts of the resultant equation must be zero. So we get:
              $$2x^2-2y^2-2x-4y-3=0$$
              and
              $$-2y-4x=0 Rightarrow y=-2x$$
              Substituting $y=-2x$ into the first real equation we get
              $$2x^2-8x^2-2x+8x-3=0$$
              or$$-6x^2+6x-3=0$$
              or $$x^2-x+0.5=0$$
              so $$x=frac{1pm i}{2}$$ which is not a real number. So there will not be any real $x$ and $y$ satisfying the complex equation. Therefore no complex $z$ satisfies the equation.







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              share|cite|improve this answer










              answered Nov 19 at 19:44









              Arash Rashidi

              538




              538






















                  up vote
                  1
                  down vote













                  $z=frac{-3+6i}{10}$ is not a solution to this problem. Let's make sure.



                  $z-1=frac{-13+6i}{10}$,



                  $bar{z}-2i=frac{-3-6i}{10}-frac{20i}{10} =frac{-3-26i}{10}$



                  $$(z-1)^2=frac{(-13+6i)^2}{100}=frac{169 - 156i-36}{100}=frac{133-156i}{100}$$



                  $$(bar{z}-2i)^2 =frac{(-3-26i)^2}{100}=frac{9-156i-676}{100}=frac{156i-667}{100}$$



                  So therefore $(z-1)^2+(bar{z}-2i)^2=-5.34$



                  In fact, there are not solutions to this equation.



                  It's not that hard to see if you apply $a^2+b^2=(a+bi)(a-bi)$
                  Taking $a=z-1$ and $b=bar{z}-2i implies bi= bar{z}i+2$



                  $a+bi=z-1+bar{z}+2=z+bar{z}+1$ and



                  $a-bi=z-1-bar{z}-2=z-bar{z}-3$ and



                  We arrive at



                  $(z-1)^2+(bar{z}-2i)^2=(z+bar{z}i+1)(z-bar{z}i-3)=0$



                  Taking $z=a+bi implies bar{z}=a-biimplies bar{z}i=b+ai$



                  So then $z+bar{z}i=(a+b)+(a+b)i$ and $z-bar{z}i= a-b-(a-b)i$
                  What's note worthy here is that the real and the imaginary parts of these numbers are the same.



                  Then $z+bar{z}i=-1 implies a+b+(a+b)i=-1$ but then $a+b$ is simulataneously $0$ and $-1$ and likewise $z-bar{z}i=3implies a-b-(a-b)i=3$ implies that the $(a-b)$ is simultaneously $3$ and $0$.



                  So we conclude: No solutions.






                  share|cite|improve this answer



























                    up vote
                    1
                    down vote













                    $z=frac{-3+6i}{10}$ is not a solution to this problem. Let's make sure.



                    $z-1=frac{-13+6i}{10}$,



                    $bar{z}-2i=frac{-3-6i}{10}-frac{20i}{10} =frac{-3-26i}{10}$



                    $$(z-1)^2=frac{(-13+6i)^2}{100}=frac{169 - 156i-36}{100}=frac{133-156i}{100}$$



                    $$(bar{z}-2i)^2 =frac{(-3-26i)^2}{100}=frac{9-156i-676}{100}=frac{156i-667}{100}$$



                    So therefore $(z-1)^2+(bar{z}-2i)^2=-5.34$



                    In fact, there are not solutions to this equation.



                    It's not that hard to see if you apply $a^2+b^2=(a+bi)(a-bi)$
                    Taking $a=z-1$ and $b=bar{z}-2i implies bi= bar{z}i+2$



                    $a+bi=z-1+bar{z}+2=z+bar{z}+1$ and



                    $a-bi=z-1-bar{z}-2=z-bar{z}-3$ and



                    We arrive at



                    $(z-1)^2+(bar{z}-2i)^2=(z+bar{z}i+1)(z-bar{z}i-3)=0$



                    Taking $z=a+bi implies bar{z}=a-biimplies bar{z}i=b+ai$



                    So then $z+bar{z}i=(a+b)+(a+b)i$ and $z-bar{z}i= a-b-(a-b)i$
                    What's note worthy here is that the real and the imaginary parts of these numbers are the same.



                    Then $z+bar{z}i=-1 implies a+b+(a+b)i=-1$ but then $a+b$ is simulataneously $0$ and $-1$ and likewise $z-bar{z}i=3implies a-b-(a-b)i=3$ implies that the $(a-b)$ is simultaneously $3$ and $0$.



                    So we conclude: No solutions.






                    share|cite|improve this answer

























                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      $z=frac{-3+6i}{10}$ is not a solution to this problem. Let's make sure.



                      $z-1=frac{-13+6i}{10}$,



                      $bar{z}-2i=frac{-3-6i}{10}-frac{20i}{10} =frac{-3-26i}{10}$



                      $$(z-1)^2=frac{(-13+6i)^2}{100}=frac{169 - 156i-36}{100}=frac{133-156i}{100}$$



                      $$(bar{z}-2i)^2 =frac{(-3-26i)^2}{100}=frac{9-156i-676}{100}=frac{156i-667}{100}$$



                      So therefore $(z-1)^2+(bar{z}-2i)^2=-5.34$



                      In fact, there are not solutions to this equation.



                      It's not that hard to see if you apply $a^2+b^2=(a+bi)(a-bi)$
                      Taking $a=z-1$ and $b=bar{z}-2i implies bi= bar{z}i+2$



                      $a+bi=z-1+bar{z}+2=z+bar{z}+1$ and



                      $a-bi=z-1-bar{z}-2=z-bar{z}-3$ and



                      We arrive at



                      $(z-1)^2+(bar{z}-2i)^2=(z+bar{z}i+1)(z-bar{z}i-3)=0$



                      Taking $z=a+bi implies bar{z}=a-biimplies bar{z}i=b+ai$



                      So then $z+bar{z}i=(a+b)+(a+b)i$ and $z-bar{z}i= a-b-(a-b)i$
                      What's note worthy here is that the real and the imaginary parts of these numbers are the same.



                      Then $z+bar{z}i=-1 implies a+b+(a+b)i=-1$ but then $a+b$ is simulataneously $0$ and $-1$ and likewise $z-bar{z}i=3implies a-b-(a-b)i=3$ implies that the $(a-b)$ is simultaneously $3$ and $0$.



                      So we conclude: No solutions.






                      share|cite|improve this answer














                      $z=frac{-3+6i}{10}$ is not a solution to this problem. Let's make sure.



                      $z-1=frac{-13+6i}{10}$,



                      $bar{z}-2i=frac{-3-6i}{10}-frac{20i}{10} =frac{-3-26i}{10}$



                      $$(z-1)^2=frac{(-13+6i)^2}{100}=frac{169 - 156i-36}{100}=frac{133-156i}{100}$$



                      $$(bar{z}-2i)^2 =frac{(-3-26i)^2}{100}=frac{9-156i-676}{100}=frac{156i-667}{100}$$



                      So therefore $(z-1)^2+(bar{z}-2i)^2=-5.34$



                      In fact, there are not solutions to this equation.



                      It's not that hard to see if you apply $a^2+b^2=(a+bi)(a-bi)$
                      Taking $a=z-1$ and $b=bar{z}-2i implies bi= bar{z}i+2$



                      $a+bi=z-1+bar{z}+2=z+bar{z}+1$ and



                      $a-bi=z-1-bar{z}-2=z-bar{z}-3$ and



                      We arrive at



                      $(z-1)^2+(bar{z}-2i)^2=(z+bar{z}i+1)(z-bar{z}i-3)=0$



                      Taking $z=a+bi implies bar{z}=a-biimplies bar{z}i=b+ai$



                      So then $z+bar{z}i=(a+b)+(a+b)i$ and $z-bar{z}i= a-b-(a-b)i$
                      What's note worthy here is that the real and the imaginary parts of these numbers are the same.



                      Then $z+bar{z}i=-1 implies a+b+(a+b)i=-1$ but then $a+b$ is simulataneously $0$ and $-1$ and likewise $z-bar{z}i=3implies a-b-(a-b)i=3$ implies that the $(a-b)$ is simultaneously $3$ and $0$.



                      So we conclude: No solutions.







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                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Nov 19 at 20:03

























                      answered Nov 19 at 19:21









                      Mason

                      1,8191427




                      1,8191427






















                          up vote
                          0
                          down vote













                          Alternative way to arrive to the conclusion that the equation has no solutions:



                          Write $;z=a+biimpliesoverline z=a-bi;$ , and the given equation is



                          $$(a-1+bi)^2+(a-(b+2)i)^2=0iff$$



                          $$(a-1)^2-b^2+2(a-1)bi+a^2-(b+2)^2-2a(b+2)i=0$$



                          and compare now real and imaginary parts:



                          $$begin{cases}(a-1)^2-b^2+a^2-(b+2)^2=0iff2a^2-2b^2-2a-4b-3=0\
                          2ab-2b-2ab-4a=0iff b=-2aend{cases};;;implies$$



                          $$2a^2-8a^2-2a+8a-3=0iff 6a^2-6a+3=0$$



                          adn the last equation has no real solution...as it should if there was a solution.






                          share|cite|improve this answer

























                            up vote
                            0
                            down vote













                            Alternative way to arrive to the conclusion that the equation has no solutions:



                            Write $;z=a+biimpliesoverline z=a-bi;$ , and the given equation is



                            $$(a-1+bi)^2+(a-(b+2)i)^2=0iff$$



                            $$(a-1)^2-b^2+2(a-1)bi+a^2-(b+2)^2-2a(b+2)i=0$$



                            and compare now real and imaginary parts:



                            $$begin{cases}(a-1)^2-b^2+a^2-(b+2)^2=0iff2a^2-2b^2-2a-4b-3=0\
                            2ab-2b-2ab-4a=0iff b=-2aend{cases};;;implies$$



                            $$2a^2-8a^2-2a+8a-3=0iff 6a^2-6a+3=0$$



                            adn the last equation has no real solution...as it should if there was a solution.






                            share|cite|improve this answer























                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              Alternative way to arrive to the conclusion that the equation has no solutions:



                              Write $;z=a+biimpliesoverline z=a-bi;$ , and the given equation is



                              $$(a-1+bi)^2+(a-(b+2)i)^2=0iff$$



                              $$(a-1)^2-b^2+2(a-1)bi+a^2-(b+2)^2-2a(b+2)i=0$$



                              and compare now real and imaginary parts:



                              $$begin{cases}(a-1)^2-b^2+a^2-(b+2)^2=0iff2a^2-2b^2-2a-4b-3=0\
                              2ab-2b-2ab-4a=0iff b=-2aend{cases};;;implies$$



                              $$2a^2-8a^2-2a+8a-3=0iff 6a^2-6a+3=0$$



                              adn the last equation has no real solution...as it should if there was a solution.






                              share|cite|improve this answer












                              Alternative way to arrive to the conclusion that the equation has no solutions:



                              Write $;z=a+biimpliesoverline z=a-bi;$ , and the given equation is



                              $$(a-1+bi)^2+(a-(b+2)i)^2=0iff$$



                              $$(a-1)^2-b^2+2(a-1)bi+a^2-(b+2)^2-2a(b+2)i=0$$



                              and compare now real and imaginary parts:



                              $$begin{cases}(a-1)^2-b^2+a^2-(b+2)^2=0iff2a^2-2b^2-2a-4b-3=0\
                              2ab-2b-2ab-4a=0iff b=-2aend{cases};;;implies$$



                              $$2a^2-8a^2-2a+8a-3=0iff 6a^2-6a+3=0$$



                              adn the last equation has no real solution...as it should if there was a solution.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 19 at 20:53









                              DonAntonio

                              176k1491224




                              176k1491224






















                                  up vote
                                  0
                                  down vote













                                  Late answer but I think worth mentioning it:



                                  You can write the expression as difference of squares and factor it into two factors which can never be $0$:
                                  $$begin{eqnarray*}(z-1)^2+(bar{z}-2i)^2 & = & (z-1)^2-i^2(bar{z}-2i)^2 \
                                  & = & (z+1 + ibar z)(z-3 - ibar z) \
                                  & = & 0
                                  end{eqnarray*}$$

                                  With $z = a+ib$ you get




                                  • $z+1 + ibar z = a+b+1 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$

                                  • $z-3 - ibar z =a+b-3 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$






                                  share|cite|improve this answer

























                                    up vote
                                    0
                                    down vote













                                    Late answer but I think worth mentioning it:



                                    You can write the expression as difference of squares and factor it into two factors which can never be $0$:
                                    $$begin{eqnarray*}(z-1)^2+(bar{z}-2i)^2 & = & (z-1)^2-i^2(bar{z}-2i)^2 \
                                    & = & (z+1 + ibar z)(z-3 - ibar z) \
                                    & = & 0
                                    end{eqnarray*}$$

                                    With $z = a+ib$ you get




                                    • $z+1 + ibar z = a+b+1 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$

                                    • $z-3 - ibar z =a+b-3 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$






                                    share|cite|improve this answer























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      Late answer but I think worth mentioning it:



                                      You can write the expression as difference of squares and factor it into two factors which can never be $0$:
                                      $$begin{eqnarray*}(z-1)^2+(bar{z}-2i)^2 & = & (z-1)^2-i^2(bar{z}-2i)^2 \
                                      & = & (z+1 + ibar z)(z-3 - ibar z) \
                                      & = & 0
                                      end{eqnarray*}$$

                                      With $z = a+ib$ you get




                                      • $z+1 + ibar z = a+b+1 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$

                                      • $z-3 - ibar z =a+b-3 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$






                                      share|cite|improve this answer












                                      Late answer but I think worth mentioning it:



                                      You can write the expression as difference of squares and factor it into two factors which can never be $0$:
                                      $$begin{eqnarray*}(z-1)^2+(bar{z}-2i)^2 & = & (z-1)^2-i^2(bar{z}-2i)^2 \
                                      & = & (z+1 + ibar z)(z-3 - ibar z) \
                                      & = & 0
                                      end{eqnarray*}$$

                                      With $z = a+ib$ you get




                                      • $z+1 + ibar z = a+b+1 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$

                                      • $z-3 - ibar z =a+b-3 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 20 at 4:21









                                      trancelocation

                                      8,5721520




                                      8,5721520






























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