Solving an equation involving complex conjugates












6














I have the following question and cannot seem to overcome how to contend with equations using $z$ and $bar z$ together. For example, the below problem:




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.
















  • 3




    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
    Dec 28 at 2:29
















6














I have the following question and cannot seem to overcome how to contend with equations using $z$ and $bar z$ together. For example, the below problem:




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.
















  • 3




    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
    Dec 28 at 2:29














6












6








6







I have the following question and cannot seem to overcome how to contend with equations using $z$ and $bar z$ together. For example, the below problem:




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 contend with equations using $z$ and $bar z$ together. For example, the below problem:




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






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.











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.









share|cite|improve this question




share|cite|improve this question








edited Dec 28 at 8:53









Eevee Trainer

4,4031632




4,4031632






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.









asked Dec 28 at 0:29









Laura Salas

333




333




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.





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.






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








  • 3




    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
    Dec 28 at 2:29














  • 3




    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
    Dec 28 at 2:29








3




3




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
Dec 28 at 2:29




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
Dec 28 at 2:29










4 Answers
4






active

oldest

votes


















15














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 everyone just gave the answers. Not good, and not what math is about, you get me?)






share|cite|improve this answer



















  • 1




    Then, we get a system of equations by equating real and imaginary parts!” OH ! This is so cool, didn’t know this could be done, but it does make sense. That’s what I was missing, thank you! As for the way you answered my question without exactly giving me the answer, that’s really what I was trying to get, an explanation and a line of reasoning so that I could then achieve it on my own. I like the mentality on this forum a lot so far. Have a good day :)
    – Laura Salas
    2 days ago





















16














Another approach is to take the complex conjugate of your equation:
$$3overline z-iz=4-i.$$
You now have two equations for $z$ and $overline z$. Now eliminate $overline z$
from them and solve for $z$.






share|cite|improve this answer





















  • !! Also works, thank you. This is a shorter way to do it.
    – Laura Salas
    2 days ago



















3














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





























    3














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



    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
      Dec 28 at 0:45






    • 1




      @LiveFreeorπHard That was a typo. I had used $3a+3ib$ in the next step. Thanks anyway.
      – Kavi Rama Murthy
      Dec 28 at 5:25











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });






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










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054472%2fsolving-an-equation-involving-complex-conjugates%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    15














    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 everyone just gave the answers. Not good, and not what math is about, you get me?)






    share|cite|improve this answer



















    • 1




      Then, we get a system of equations by equating real and imaginary parts!” OH ! This is so cool, didn’t know this could be done, but it does make sense. That’s what I was missing, thank you! As for the way you answered my question without exactly giving me the answer, that’s really what I was trying to get, an explanation and a line of reasoning so that I could then achieve it on my own. I like the mentality on this forum a lot so far. Have a good day :)
      – Laura Salas
      2 days ago


















    15














    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 everyone just gave the answers. Not good, and not what math is about, you get me?)






    share|cite|improve this answer



















    • 1




      Then, we get a system of equations by equating real and imaginary parts!” OH ! This is so cool, didn’t know this could be done, but it does make sense. That’s what I was missing, thank you! As for the way you answered my question without exactly giving me the answer, that’s really what I was trying to get, an explanation and a line of reasoning so that I could then achieve it on my own. I like the mentality on this forum a lot so far. Have a good day :)
      – Laura Salas
      2 days ago
















    15












    15








    15






    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 everyone 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 everyone 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








    edited Dec 28 at 8:52

























    answered Dec 28 at 0:40









    Eevee Trainer

    4,4031632




    4,4031632








    • 1




      Then, we get a system of equations by equating real and imaginary parts!” OH ! This is so cool, didn’t know this could be done, but it does make sense. That’s what I was missing, thank you! As for the way you answered my question without exactly giving me the answer, that’s really what I was trying to get, an explanation and a line of reasoning so that I could then achieve it on my own. I like the mentality on this forum a lot so far. Have a good day :)
      – Laura Salas
      2 days ago
















    • 1




      Then, we get a system of equations by equating real and imaginary parts!” OH ! This is so cool, didn’t know this could be done, but it does make sense. That’s what I was missing, thank you! As for the way you answered my question without exactly giving me the answer, that’s really what I was trying to get, an explanation and a line of reasoning so that I could then achieve it on my own. I like the mentality on this forum a lot so far. Have a good day :)
      – Laura Salas
      2 days ago










    1




    1




    Then, we get a system of equations by equating real and imaginary parts!” OH ! This is so cool, didn’t know this could be done, but it does make sense. That’s what I was missing, thank you! As for the way you answered my question without exactly giving me the answer, that’s really what I was trying to get, an explanation and a line of reasoning so that I could then achieve it on my own. I like the mentality on this forum a lot so far. Have a good day :)
    – Laura Salas
    2 days ago






    Then, we get a system of equations by equating real and imaginary parts!” OH ! This is so cool, didn’t know this could be done, but it does make sense. That’s what I was missing, thank you! As for the way you answered my question without exactly giving me the answer, that’s really what I was trying to get, an explanation and a line of reasoning so that I could then achieve it on my own. I like the mentality on this forum a lot so far. Have a good day :)
    – Laura Salas
    2 days ago













    16














    Another approach is to take the complex conjugate of your equation:
    $$3overline z-iz=4-i.$$
    You now have two equations for $z$ and $overline z$. Now eliminate $overline z$
    from them and solve for $z$.






    share|cite|improve this answer





















    • !! Also works, thank you. This is a shorter way to do it.
      – Laura Salas
      2 days ago
















    16














    Another approach is to take the complex conjugate of your equation:
    $$3overline z-iz=4-i.$$
    You now have two equations for $z$ and $overline z$. Now eliminate $overline z$
    from them and solve for $z$.






    share|cite|improve this answer





















    • !! Also works, thank you. This is a shorter way to do it.
      – Laura Salas
      2 days ago














    16












    16








    16






    Another approach is to take the complex conjugate of your equation:
    $$3overline z-iz=4-i.$$
    You now have two equations for $z$ and $overline z$. Now eliminate $overline z$
    from them and solve for $z$.






    share|cite|improve this answer












    Another approach is to take the complex conjugate of your equation:
    $$3overline z-iz=4-i.$$
    You now have two equations for $z$ and $overline z$. Now eliminate $overline z$
    from them and solve for $z$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 28 at 3:37









    Lord Shark the Unknown

    101k958132




    101k958132












    • !! Also works, thank you. This is a shorter way to do it.
      – Laura Salas
      2 days ago


















    • !! Also works, thank you. This is a shorter way to do it.
      – Laura Salas
      2 days ago
















    !! Also works, thank you. This is a shorter way to do it.
    – Laura Salas
    2 days ago




    !! Also works, thank you. This is a shorter way to do it.
    – Laura Salas
    2 days ago











    3














    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


























      3














      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
























        3












        3








        3






        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 Dec 28 at 0:52









        Rhys Hughes

        4,8041327




        4,8041327























            3














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



            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
              Dec 28 at 0:45






            • 1




              @LiveFreeorπHard That was a typo. I had used $3a+3ib$ in the next step. Thanks anyway.
              – Kavi Rama Murthy
              Dec 28 at 5:25
















            3














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



            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
              Dec 28 at 0:45






            • 1




              @LiveFreeorπHard That was a typo. I had used $3a+3ib$ in the next step. Thanks anyway.
              – Kavi Rama Murthy
              Dec 28 at 5:25














            3












            3








            3






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



            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 $$(3a+3ib)+i(a-ib)=4+i$$



            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 Dec 28 at 4:58

























            answered Dec 28 at 0:42









            Kavi Rama Murthy

            50.1k31854




            50.1k31854








            • 1




              It’s $3a + 3ib$ in the first bracket of your first equation.
              – Live Free or π Hard
              Dec 28 at 0:45






            • 1




              @LiveFreeorπHard That was a typo. I had used $3a+3ib$ in the next step. Thanks anyway.
              – Kavi Rama Murthy
              Dec 28 at 5:25














            • 1




              It’s $3a + 3ib$ in the first bracket of your first equation.
              – Live Free or π Hard
              Dec 28 at 0:45






            • 1




              @LiveFreeorπHard That was a typo. I had used $3a+3ib$ in the next step. Thanks anyway.
              – Kavi Rama Murthy
              Dec 28 at 5:25








            1




            1




            It’s $3a + 3ib$ in the first bracket of your first equation.
            – Live Free or π Hard
            Dec 28 at 0:45




            It’s $3a + 3ib$ in the first bracket of your first equation.
            – Live Free or π Hard
            Dec 28 at 0:45




            1




            1




            @LiveFreeorπHard That was a typo. I had used $3a+3ib$ in the next step. Thanks anyway.
            – Kavi Rama Murthy
            Dec 28 at 5:25




            @LiveFreeorπHard That was a typo. I had used $3a+3ib$ in the next step. Thanks anyway.
            – Kavi Rama Murthy
            Dec 28 at 5:25










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










            draft saved

            draft discarded


















            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.
















            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054472%2fsolving-an-equation-involving-complex-conjugates%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Human spaceflight

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            張江高科駅