Solving an equation involving complex conjugates 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.
 A: 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 $z=\frac {11} 8-i\frac 1 8$.
A: 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:
$$3a+b=4$$
$$a+3b=1$$
A: 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 $z$.
A: 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?)
A: In Mathematica or Wolfram Alpha, the code below returns $z=\frac{11}{8}-\frac{i}{8}.$
Solve[3 z + I Conjugate[z] == 4 + I, z]

