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How to solve $|z^2-1|<|z|^2$ where $z$ is a complex number? I have tried it both with cartesian and polar coordinates but did not get a solution.

I got that far: $z=x+yi$ and then I got: $$\pm x >(\frac{y^2+0.5}{1+4y^2})^{0.5}$$ but I don't know how to visualise that in the coordinate system.

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That is equivalent to $$|z^2-1|<|z^2|$$ This means that $z^2$ is at a shorter distance from $1$ than from $0$. Then $Re(z^2)>1/2$.

Now, write $z=x+iy$, thus, $z^2=(x^2-y^2)+2xyi$. The former inequality becomes $$x^2-y^2>\frac12$$

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  • $\begingroup$ No sorry, I can't. Can you help me? $\endgroup$ – user50224 Nov 21 '14 at 14:03
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With $\;z=x+iy\;$ :

$$|z^2-1|<|z|^2\iff |(x^2-y^2-1)+2xyi|<|x+iy|^2\iff$$

$$\sqrt{(x^2-y^2-1)^2+4x^2y^2}<x^2+y^2\iff(x^2-y^2-1)^2+4x^2y^2<(x^2+y^2)^2\iff$$

$$\color{red}{x^4}-\color{green}{2x^2y^2}+\color{blue}{y^4}-2x^2+2y^2+1+\color{green}{4x^2y^2}<\color{red}{x^4}+\color{green}{2x^2y^2}+\color{blue}{y^4}\iff$$

$$-2x^2+2y^2+1<0\iff x^2-y^2>\frac12$$

This already looks as the exterior of a rather simple hyperbola...

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  • $\begingroup$ This looks good but why is the $2xyi$ from your first line, after you squared it, not $-4x^2 y^2$? $\endgroup$ – user50224 Nov 21 '14 at 13:57
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    $\begingroup$ @user50224 Definitiobn of module (=absolute value) of complex number: if $\;x,y\in\Bbb R\;$ , then $$|x+iy|:=\sqrt{x^2+y^2}$$ $\endgroup$ – Timbuc Nov 21 '14 at 14:05

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