Determine the values $z = a+ib \in \mathbb{C}$ such that $Im((2-i)z) \leq 1 \\ $

Question

Determine the values $z = a+ib \in \mathbb{C}$ such that $Im((2-i)z) \leq 1 \\ $

Here's what I've got so far:

$Im((2-i)z) \leq 1 \\ $

$(2-i)z = 2z-iz \\ $

$=2(a+bi) - (a+bi)i = (2a+b)+i(2b-a)\\ $

$So \: Im((2a+b)+i(2b-a)) = 2b-a \\ $

$\Rightarrow 2b-a \leq 1+0i \\ $

But I'm not sure where to go from here, or if this working is correct so far. Any help would be great!

Answer 1

There is no need to introduce $i$ at the end. You're after those $a+bi$ ($a,b\in\mathbb R$) such that $\operatorname{Im}\bigl((2-i)(a+bi)\bigr)\leqslant1$ and you proved that this is equivalent to the assertion that $2b-a\leqslant1$. So, the answer is:$$\{a+bi\in\mathbb C\mid a,b\in\mathbb R\text{ and }2b-a\leqslant1\}.$$You can also say that the answer is$$\{z\in\mathbb C\mid2\operatorname{Im}(z)-\operatorname{Re}(z)\leqslant1\}.$$

Answer 2

You can write $z=a+ib$ and so: $$\Im((2-i)\cdot(a+ib))=\Im((2a+b)+i(2b-a))=2b+a$$ Using the condition $2b+a$, you can deduce that the locus is: $$L=\{w\in C|\Re(w)+2\Im(w)\geq0\}$$