Find $z$ - complex numbers

Question

Find $z$ if

$\frac{z}{z+2} = 2 - i$

To solve this question, I used $a+bi$ in place of $z$. I finally arrived at the answer $z=-3-i$. This has been the case for many complex number questions. I usually substitute $a+bi$ and compare both sides of the equation. Is there any other more elegant or "smarter" method to do this question?

Answer 1

I would have done it as follows:\begin{align}\frac z{z+2}=2-i&\iff z=(2-i)(z+2)\\&\iff z\bigl(1-(2-i)\bigr)=4-2i\\&\iff z=\frac{4-2i}{-1+i}=-3-i.\end{align}

Answer 2

Hint: Multiplying by $z+2$ we get $$z=(2-i)(z+2)$$ so $$z=2z-iz+4-2i$$ or $$z(i-1)=4-2i$$ Can you finish?

Answer 3

For example we can write

$$\frac{z}{z+2}=2-i \implies\frac{z+2-2}{z+2}=2-i \implies 1-\frac{2}{z+2}=2-i$$

$$ \implies i-1 = \frac{2}{z+2} \implies z+2 = \frac{2}{ i-1} \implies z = \frac{2}{ i-1}-2$$

Then simplify the right-hand-side.

Answer 4

$\dfrac{z+2}z=\dfrac1{2-i}$

$\dfrac2z=\dfrac{i-1}{2-i}$

$-\dfrac z2=\dfrac{2-i}{1-i}$

Rationalize the denominator