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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: $$3z+i\bar z=4+i$$

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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    $\begingroup$ 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? $\endgroup$ – Eric Towers Dec 28 '18 at 2:29
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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

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

Then, making our substitutions...

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

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?)

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    $\begingroup$ 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 :) $\endgroup$ – Laura Salas Dec 29 '18 at 21:26
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Another approach is to take the complex conjugate of your equation: $$3\overline z-iz=4-i.$$ You now have two equations for $z$ and $\overline z$. Now eliminate $\overline z$ from them and solve for $z$.

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  • $\begingroup$ !! Also works, thank you. This is a shorter way to do it. $\endgroup$ – Laura Salas Dec 29 '18 at 21:27
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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:

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

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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 $z=\frac {11} 8-i\frac 1 8$.

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    $\begingroup$ It’s $3a + 3ib$ in the first bracket of your first equation. $\endgroup$ – Live Free or π Hard Dec 28 '18 at 0:45
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    $\begingroup$ @LiveFreeorπHard That was a typo. I had used $3a+3ib$ in the next step. Thanks anyway. $\endgroup$ – Kavi Rama Murthy Dec 28 '18 at 5:25

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