# Complex numbers: With conjugate

I've just started calculating complex numbers (last time I calculated with complex numbers was in high school) and I've already got stuck at this exercise:

$$3z-i\bar z = 7-5i$$

where $\bar z$ is the conjugate of z.

What I've tested so far is to set $z=x+yi$

and with further calculations I've reached this equation

$$3(x+yi)-i(x-yi)=7-5i \implies 3x+3yi-xi+yi^2=7-5i$$

The result should be $z=2-i$.

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You can format your answers at MSE with $\LaTeX!$. Read basic manual at this meta post. – FrenzY DT. Dec 4 '12 at 16:06

You have as your last equation $$3x + 3yi - xi + yi^2 = 7-5i$$ now $i^2 = -1$, so we have $$3x-y + (3y - x)i = 7-5i$$ Now as $x$ and $y$ are real, we must have (as complex numbers are equal iff both real and imaginary parts are) $$3x-y = 7 \land 3y - x = -5$$ This is a linear system for $x$ and $y$ which has $x = 2$ and $y = -1$ as its solution. Hence $z = x+yi = 2-i$.
If we know $i^2=-1$ and $x+iy=a+ib\iff x=a,y=b$
$$3x-y+i(3y-x)=7-5i$$
So, $3x-y=7$ and $3y-x=-5$