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I have the equation: $$3z-\bar{z}=2-3i$$

First I write this as: $$3(x+yi)-(x-yi) = 2-3i$$ $$3x-3yi-x+yi = 2-3i$$ $$2x+4yi = 2-3i$$

Now the following must be true: $$2x = 2\quad\mbox{and}\quad 4y = -3$$

So $x = 1$, and $y=\textstyle\dfrac{-3i}{4}$

Thus $$z=1-\textstyle\dfrac{3i}{4}$$

Is this correct?

JMP
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user140878
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    Yes, that's correct. You are justified in equating the real and imaginary parts on the two sides of the equation. – Mark Dec 09 '14 at 15:27
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    It is correct. Only note that you have a typo in the third line. $=3x-3yi-\cdots$ should read $=3x+3yi-\cdots$ – mfl Dec 09 '14 at 15:35

4 Answers4

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Right idea, two small errors.

In line 3 the first minus sign should be a plus sign: $$3x+3yi-x+yi=2-3i$$ but that was just a typo. Farther down where you wrote $$y=\frac{-3i}4$$ you meant $$y=\frac{-3}4.$$

bof
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It is correct, but you can do it in a slicker way: conjugate and find $$ 3\bar{z}-z=2+3i $$ But the original equation says $\bar{z}=3z-2+3i$, so, substituting gives $$ 9z-6+9i-z=2+3i $$ that becomes $$ 8z=8-6i $$ that is, $$ z=1-\frac{3}{4}i $$

egreg
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Yes, it is correct. You can check you work by plugging the $z$ you found back into the original equation. $$3\left(1-\frac{3i}{4}\right)-\left(1+\frac{3i}{4}\right)=2-3i\\3-\frac{9i}{4}-1-\frac{3i}{4}=2-3i\\2-\frac{12i}{4}=2-3i\\2-3i=2-3i.$$ So your $z=1-\frac{3i}{4}$ is correct.

homegrown
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Here are the steps $$ 3z-\overline{z}=2-3i $$ $$ 3(x+yi)-(x-yi)=2-3i $$ $$ 3x+3yi-x+yi=2-3i $$ $$ 2x+4yi=2-3i $$ Therefore, $$ x=1\quad\mbox{and}\quad y=-\frac34 $$ Which implies that $$ z=1-\frac34i$$

k170
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