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I'm trying to find the solution of this inequality but I don't know what to do next: $$(|\overline{z} - i|-2) \le 0 \rightarrow (|a-i(b-1)| -2) \le 0 \rightarrow \sqrt{a^2 + b^2 + 2b +1} \le 2 \rightarrow a^2 + b^2 + 2b \le 1$$ With $ z\in \mathbb{C}$. As it can be seen I have an inequality in 2 unknowns. How do I solve this type of inequalities?

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    $\begingroup$ You should be familiar with $|z-z_0|\leq c$ as being disks of radius $c$ centered at $z_0$. There are only a few small details different about this and your question. $\endgroup$ – JMoravitz Nov 19 '16 at 1:02
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    $\begingroup$ Also, as an aside, your final step seems incorrect, $\sqrt{a^2+b^2+2b+1}\leq 2$, by squaring both sides, we have then instead $a^2+b^2+2b+1\leq 4$. Moving the $1$ to the other side, that would be an upper bound of three, not of one. $\endgroup$ – JMoravitz Nov 19 '16 at 1:05
  • $\begingroup$ @JMoravitz thanks for the last remark, though I'm still trying to figure out the first. I know what disks of radius c are but I can't make that equation meet the requirements for one. As you can see if I expend $\overline{z}$ it will just make the equation more compicated and I won't arrive at the form: $|z-z_0| \le c$ $\endgroup$ – Tay123 Nov 19 '16 at 1:10
  • $\begingroup$ Tell me... is $|z|=|\overline{z}|$? $\endgroup$ – JMoravitz Nov 19 '16 at 1:11
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    $\begingroup$ @JMoravitz now I get it; you did $ |z|=|\overline{z}| \rightarrow |\overline{z}-i| = |\overline{z+i}| \rightarrow |\overline{z+i}| = |z+i| \rightarrow |\overline{z}-i| = |z+i| $ Thanks for the help. If you want you can answer so I can vote you for the help given... if not I'll just delete the question (if it's not against the rules, obviously). $\endgroup$ – Tay123 Nov 19 '16 at 1:20
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Since it has been figured out in the comments, there are a few things to note here:

  • The inequality $|z-z_0|\leq c$ describes a disc of radius $c$ centered at $z_0$

  • More generally, the inequality $\frac{(x-x_0)^2}{a}+\frac{(y-y_0)^2}{b}\leq 1$ in $\Bbb R^2$ describes a filled ellipse with center $(x_0,y_0)$ with corresponding axis lengths $a$ and $b$

  • $|z|=|\overline{z}|$

We note then:

$0\geq |\overline{z}-i|-2=|\overline{(\overline{z}-i)}|-2=|z+i|-2$

This implies $|z-(-i)|\leq 2$ and so the inequality describes the disc of radius $2$ centered at $-i$.

One can be even more general for equations of ellipses whose axes are not necessarily perpendicular to the usual $x$ and $y$ axes. Read more about quadratic forms on wikipedia or in standard pre-calculus or linear algebra textbooks.

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  • $\begingroup$ Thanks for the detailed explanation. $\endgroup$ – Tay123 Nov 19 '16 at 1:42

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