Solving an equation involving complex conjugates












3














I have the following question and cannot seem to overcome how to mix z and conjugate z together.




Find the value of $z in Bbb C$ that verifies the equation:
$$3z+ibar z=4+i$$




For other operations that didn't include mixing $z$ and $bar z$, I was able to manage by "isolating" $z$ on one side of the equation and finding the real and imaginary parts of the complex numbers (sorry if I'm not using the right terms, it's my first linear algebra course)



I tried with wolfram and it didn't really help.



PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.










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  • Have you tried picking a basis for $mathbb{C}$, writing $z$ as a generic vector in that space using that basis, then solving for the components of $z$? I mean, ..., this is a linear algebra problem; why not "do the linear algebra thing" to it?
    – Eric Towers
    1 hour ago
















3














I have the following question and cannot seem to overcome how to mix z and conjugate z together.




Find the value of $z in Bbb C$ that verifies the equation:
$$3z+ibar z=4+i$$




For other operations that didn't include mixing $z$ and $bar z$, I was able to manage by "isolating" $z$ on one side of the equation and finding the real and imaginary parts of the complex numbers (sorry if I'm not using the right terms, it's my first linear algebra course)



I tried with wolfram and it didn't really help.



PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.










share|cite|improve this question









New contributor




Laura Salas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Have you tried picking a basis for $mathbb{C}$, writing $z$ as a generic vector in that space using that basis, then solving for the components of $z$? I mean, ..., this is a linear algebra problem; why not "do the linear algebra thing" to it?
    – Eric Towers
    1 hour ago














3












3








3







I have the following question and cannot seem to overcome how to mix z and conjugate z together.




Find the value of $z in Bbb C$ that verifies the equation:
$$3z+ibar z=4+i$$




For other operations that didn't include mixing $z$ and $bar z$, I was able to manage by "isolating" $z$ on one side of the equation and finding the real and imaginary parts of the complex numbers (sorry if I'm not using the right terms, it's my first linear algebra course)



I tried with wolfram and it didn't really help.



PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.










share|cite|improve this question









New contributor




Laura Salas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I have the following question and cannot seem to overcome how to mix z and conjugate z together.




Find the value of $z in Bbb C$ that verifies the equation:
$$3z+ibar z=4+i$$




For other operations that didn't include mixing $z$ and $bar z$, I was able to manage by "isolating" $z$ on one side of the equation and finding the real and imaginary parts of the complex numbers (sorry if I'm not using the right terms, it's my first linear algebra course)



I tried with wolfram and it didn't really help.



PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.







linear-algebra complex-numbers






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Check out our Code of Conduct.











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edited 17 mins ago









Eevee Trainer

3,880428




3,880428






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asked 3 hours ago









Laura Salas

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Laura Salas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Laura Salas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Have you tried picking a basis for $mathbb{C}$, writing $z$ as a generic vector in that space using that basis, then solving for the components of $z$? I mean, ..., this is a linear algebra problem; why not "do the linear algebra thing" to it?
    – Eric Towers
    1 hour ago


















  • Have you tried picking a basis for $mathbb{C}$, writing $z$ as a generic vector in that space using that basis, then solving for the components of $z$? I mean, ..., this is a linear algebra problem; why not "do the linear algebra thing" to it?
    – Eric Towers
    1 hour ago
















Have you tried picking a basis for $mathbb{C}$, writing $z$ as a generic vector in that space using that basis, then solving for the components of $z$? I mean, ..., this is a linear algebra problem; why not "do the linear algebra thing" to it?
– Eric Towers
1 hour ago




Have you tried picking a basis for $mathbb{C}$, writing $z$ as a generic vector in that space using that basis, then solving for the components of $z$? I mean, ..., this is a linear algebra problem; why not "do the linear algebra thing" to it?
– Eric Towers
1 hour ago










3 Answers
3






active

oldest

votes


















2














Hint:



Let $z = x + iy$, for $x,y in mathbb{R}$. Consequently, $bar{z} = x - iy$.



Make these substitutions into your equation and isolate all of the $x$ and $y$ terms on one side, trying to make it "look" like a number in that form above (I really don't know how else to describe it, my example below will be more illustrative).



Equate the real and imaginary parts to get a system of equations in two variables ($x,y$) which you can solve get your solution.





Similar Exercise To Show What I Mean:



Let's solve for $z$ with



$$iz + 2bar{z} = 1 + 2i$$



Then, making our substitutions...



$$begin{align}
iz + 2bar{z} &= i(x + iy) + 2(x - iy) \
&= ix + i^2 y + 2x - 2iy \
&= ix - y + 2x - 2iy \
&= (2x - y) + i(x - 2y) \
end{align}$$



Thus,



$$ (2x - y) + i(x - 2y) = 1 + 2i$$



The real part of our left side is $2x-y$ and the imaginary part is $x - 2y$. On the right, the real and imaginary parts are $1$ and $2$ respectively.



Then, we get a system of equations by equating real and imaginary parts!



$$begin{align}
2x - y &= 1\
x - 2y &= 2\
end{align}$$



You can quickly show with basic algebra that $y = -1, x = 0$.



Our solution is a $z$ of the form $z = x + iy$. Thus, $z = 0 + i(-1) = -i$.





One Final Tidbit:




PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.




This forum doesn't mind helping you with homework, so long as you show you make a reasonable effort or at least have a clear understanding of the material. However, the goal is also to help you learn, so people tend to prefer nudges in the right direction if the context allows it, as opposed to just handing you the solution. (Imagine how people would abuse the site for homework if every just gave the answers. Not good, and not what math is about, you get me?)






share|cite|improve this answer





























    1














    If $z=a+bi$ then $bar z=a-bi$



    So you are solving:
    $$3(a+bi)+i(a-bi)=4+i$$
    $$to (3a+b)+(a+3b)i=4+i$$
    Hence solve the simultaneous equations:



    $$3a+b=4$$
    $$a+3b=1$$






    share|cite|improve this answer





























      0














      Let $a$ and $b$ be the real and imaginary parts of $z$. The equation becomes $$(3a+ib)+i(a-ib)=4+i$$



      Equating real and imaginary parts you get $3a+b= 4$ and $3b+a=1$. Now you should be able to discover that $a=frac {11} 8$ and $b =-frac 1 8$, so $z=frac {11} 8-ifrac 1 8$.






      share|cite|improve this answer



















      • 1




        It’s $3a + 3ib$ in the first bracket of your first equation.
        – Live Free or π Hard
        2 hours ago











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      Hint:



      Let $z = x + iy$, for $x,y in mathbb{R}$. Consequently, $bar{z} = x - iy$.



      Make these substitutions into your equation and isolate all of the $x$ and $y$ terms on one side, trying to make it "look" like a number in that form above (I really don't know how else to describe it, my example below will be more illustrative).



      Equate the real and imaginary parts to get a system of equations in two variables ($x,y$) which you can solve get your solution.





      Similar Exercise To Show What I Mean:



      Let's solve for $z$ with



      $$iz + 2bar{z} = 1 + 2i$$



      Then, making our substitutions...



      $$begin{align}
      iz + 2bar{z} &= i(x + iy) + 2(x - iy) \
      &= ix + i^2 y + 2x - 2iy \
      &= ix - y + 2x - 2iy \
      &= (2x - y) + i(x - 2y) \
      end{align}$$



      Thus,



      $$ (2x - y) + i(x - 2y) = 1 + 2i$$



      The real part of our left side is $2x-y$ and the imaginary part is $x - 2y$. On the right, the real and imaginary parts are $1$ and $2$ respectively.



      Then, we get a system of equations by equating real and imaginary parts!



      $$begin{align}
      2x - y &= 1\
      x - 2y &= 2\
      end{align}$$



      You can quickly show with basic algebra that $y = -1, x = 0$.



      Our solution is a $z$ of the form $z = x + iy$. Thus, $z = 0 + i(-1) = -i$.





      One Final Tidbit:




      PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.




      This forum doesn't mind helping you with homework, so long as you show you make a reasonable effort or at least have a clear understanding of the material. However, the goal is also to help you learn, so people tend to prefer nudges in the right direction if the context allows it, as opposed to just handing you the solution. (Imagine how people would abuse the site for homework if every just gave the answers. Not good, and not what math is about, you get me?)






      share|cite|improve this answer


























        2














        Hint:



        Let $z = x + iy$, for $x,y in mathbb{R}$. Consequently, $bar{z} = x - iy$.



        Make these substitutions into your equation and isolate all of the $x$ and $y$ terms on one side, trying to make it "look" like a number in that form above (I really don't know how else to describe it, my example below will be more illustrative).



        Equate the real and imaginary parts to get a system of equations in two variables ($x,y$) which you can solve get your solution.





        Similar Exercise To Show What I Mean:



        Let's solve for $z$ with



        $$iz + 2bar{z} = 1 + 2i$$



        Then, making our substitutions...



        $$begin{align}
        iz + 2bar{z} &= i(x + iy) + 2(x - iy) \
        &= ix + i^2 y + 2x - 2iy \
        &= ix - y + 2x - 2iy \
        &= (2x - y) + i(x - 2y) \
        end{align}$$



        Thus,



        $$ (2x - y) + i(x - 2y) = 1 + 2i$$



        The real part of our left side is $2x-y$ and the imaginary part is $x - 2y$. On the right, the real and imaginary parts are $1$ and $2$ respectively.



        Then, we get a system of equations by equating real and imaginary parts!



        $$begin{align}
        2x - y &= 1\
        x - 2y &= 2\
        end{align}$$



        You can quickly show with basic algebra that $y = -1, x = 0$.



        Our solution is a $z$ of the form $z = x + iy$. Thus, $z = 0 + i(-1) = -i$.





        One Final Tidbit:




        PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.




        This forum doesn't mind helping you with homework, so long as you show you make a reasonable effort or at least have a clear understanding of the material. However, the goal is also to help you learn, so people tend to prefer nudges in the right direction if the context allows it, as opposed to just handing you the solution. (Imagine how people would abuse the site for homework if every just gave the answers. Not good, and not what math is about, you get me?)






        share|cite|improve this answer
























          2












          2








          2






          Hint:



          Let $z = x + iy$, for $x,y in mathbb{R}$. Consequently, $bar{z} = x - iy$.



          Make these substitutions into your equation and isolate all of the $x$ and $y$ terms on one side, trying to make it "look" like a number in that form above (I really don't know how else to describe it, my example below will be more illustrative).



          Equate the real and imaginary parts to get a system of equations in two variables ($x,y$) which you can solve get your solution.





          Similar Exercise To Show What I Mean:



          Let's solve for $z$ with



          $$iz + 2bar{z} = 1 + 2i$$



          Then, making our substitutions...



          $$begin{align}
          iz + 2bar{z} &= i(x + iy) + 2(x - iy) \
          &= ix + i^2 y + 2x - 2iy \
          &= ix - y + 2x - 2iy \
          &= (2x - y) + i(x - 2y) \
          end{align}$$



          Thus,



          $$ (2x - y) + i(x - 2y) = 1 + 2i$$



          The real part of our left side is $2x-y$ and the imaginary part is $x - 2y$. On the right, the real and imaginary parts are $1$ and $2$ respectively.



          Then, we get a system of equations by equating real and imaginary parts!



          $$begin{align}
          2x - y &= 1\
          x - 2y &= 2\
          end{align}$$



          You can quickly show with basic algebra that $y = -1, x = 0$.



          Our solution is a $z$ of the form $z = x + iy$. Thus, $z = 0 + i(-1) = -i$.





          One Final Tidbit:




          PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.




          This forum doesn't mind helping you with homework, so long as you show you make a reasonable effort or at least have a clear understanding of the material. However, the goal is also to help you learn, so people tend to prefer nudges in the right direction if the context allows it, as opposed to just handing you the solution. (Imagine how people would abuse the site for homework if every just gave the answers. Not good, and not what math is about, you get me?)






          share|cite|improve this answer












          Hint:



          Let $z = x + iy$, for $x,y in mathbb{R}$. Consequently, $bar{z} = x - iy$.



          Make these substitutions into your equation and isolate all of the $x$ and $y$ terms on one side, trying to make it "look" like a number in that form above (I really don't know how else to describe it, my example below will be more illustrative).



          Equate the real and imaginary parts to get a system of equations in two variables ($x,y$) which you can solve get your solution.





          Similar Exercise To Show What I Mean:



          Let's solve for $z$ with



          $$iz + 2bar{z} = 1 + 2i$$



          Then, making our substitutions...



          $$begin{align}
          iz + 2bar{z} &= i(x + iy) + 2(x - iy) \
          &= ix + i^2 y + 2x - 2iy \
          &= ix - y + 2x - 2iy \
          &= (2x - y) + i(x - 2y) \
          end{align}$$



          Thus,



          $$ (2x - y) + i(x - 2y) = 1 + 2i$$



          The real part of our left side is $2x-y$ and the imaginary part is $x - 2y$. On the right, the real and imaginary parts are $1$ and $2$ respectively.



          Then, we get a system of equations by equating real and imaginary parts!



          $$begin{align}
          2x - y &= 1\
          x - 2y &= 2\
          end{align}$$



          You can quickly show with basic algebra that $y = -1, x = 0$.



          Our solution is a $z$ of the form $z = x + iy$. Thus, $z = 0 + i(-1) = -i$.





          One Final Tidbit:




          PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.




          This forum doesn't mind helping you with homework, so long as you show you make a reasonable effort or at least have a clear understanding of the material. However, the goal is also to help you learn, so people tend to prefer nudges in the right direction if the context allows it, as opposed to just handing you the solution. (Imagine how people would abuse the site for homework if every just gave the answers. Not good, and not what math is about, you get me?)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Eevee Trainer

          3,880428




          3,880428























              1














              If $z=a+bi$ then $bar z=a-bi$



              So you are solving:
              $$3(a+bi)+i(a-bi)=4+i$$
              $$to (3a+b)+(a+3b)i=4+i$$
              Hence solve the simultaneous equations:



              $$3a+b=4$$
              $$a+3b=1$$






              share|cite|improve this answer


























                1














                If $z=a+bi$ then $bar z=a-bi$



                So you are solving:
                $$3(a+bi)+i(a-bi)=4+i$$
                $$to (3a+b)+(a+3b)i=4+i$$
                Hence solve the simultaneous equations:



                $$3a+b=4$$
                $$a+3b=1$$






                share|cite|improve this answer
























                  1












                  1








                  1






                  If $z=a+bi$ then $bar z=a-bi$



                  So you are solving:
                  $$3(a+bi)+i(a-bi)=4+i$$
                  $$to (3a+b)+(a+3b)i=4+i$$
                  Hence solve the simultaneous equations:



                  $$3a+b=4$$
                  $$a+3b=1$$






                  share|cite|improve this answer












                  If $z=a+bi$ then $bar z=a-bi$



                  So you are solving:
                  $$3(a+bi)+i(a-bi)=4+i$$
                  $$to (3a+b)+(a+3b)i=4+i$$
                  Hence solve the simultaneous equations:



                  $$3a+b=4$$
                  $$a+3b=1$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Rhys Hughes

                  4,7511327




                  4,7511327























                      0














                      Let $a$ and $b$ be the real and imaginary parts of $z$. The equation becomes $$(3a+ib)+i(a-ib)=4+i$$



                      Equating real and imaginary parts you get $3a+b= 4$ and $3b+a=1$. Now you should be able to discover that $a=frac {11} 8$ and $b =-frac 1 8$, so $z=frac {11} 8-ifrac 1 8$.






                      share|cite|improve this answer



















                      • 1




                        It’s $3a + 3ib$ in the first bracket of your first equation.
                        – Live Free or π Hard
                        2 hours ago
















                      0














                      Let $a$ and $b$ be the real and imaginary parts of $z$. The equation becomes $$(3a+ib)+i(a-ib)=4+i$$



                      Equating real and imaginary parts you get $3a+b= 4$ and $3b+a=1$. Now you should be able to discover that $a=frac {11} 8$ and $b =-frac 1 8$, so $z=frac {11} 8-ifrac 1 8$.






                      share|cite|improve this answer



















                      • 1




                        It’s $3a + 3ib$ in the first bracket of your first equation.
                        – Live Free or π Hard
                        2 hours ago














                      0












                      0








                      0






                      Let $a$ and $b$ be the real and imaginary parts of $z$. The equation becomes $$(3a+ib)+i(a-ib)=4+i$$



                      Equating real and imaginary parts you get $3a+b= 4$ and $3b+a=1$. Now you should be able to discover that $a=frac {11} 8$ and $b =-frac 1 8$, so $z=frac {11} 8-ifrac 1 8$.






                      share|cite|improve this answer














                      Let $a$ and $b$ be the real and imaginary parts of $z$. The equation becomes $$(3a+ib)+i(a-ib)=4+i$$



                      Equating real and imaginary parts you get $3a+b= 4$ and $3b+a=1$. Now you should be able to discover that $a=frac {11} 8$ and $b =-frac 1 8$, so $z=frac {11} 8-ifrac 1 8$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 2 hours ago









                      Eevee Trainer

                      3,880428




                      3,880428










                      answered 2 hours ago









                      Kavi Rama Murthy

                      49.4k31854




                      49.4k31854








                      • 1




                        It’s $3a + 3ib$ in the first bracket of your first equation.
                        – Live Free or π Hard
                        2 hours ago














                      • 1




                        It’s $3a + 3ib$ in the first bracket of your first equation.
                        – Live Free or π Hard
                        2 hours ago








                      1




                      1




                      It’s $3a + 3ib$ in the first bracket of your first equation.
                      – Live Free or π Hard
                      2 hours ago




                      It’s $3a + 3ib$ in the first bracket of your first equation.
                      – Live Free or π Hard
                      2 hours ago










                      Laura Salas is a new contributor. Be nice, and check out our Code of Conduct.










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                      Laura Salas is a new contributor. Be nice, and check out our Code of Conduct.













                      Laura Salas is a new contributor. Be nice, and check out our Code of Conduct.












                      Laura Salas is a new contributor. Be nice, and check out our Code of Conduct.
















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