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Determine the set of points $z$ that satisfy the condition $|2z|>|1+z^2|$

I tried to redo this problem and got to this point

$|2z|>|1+z^2|$ $\Rightarrow$ $2|z|>1+|z^2|$ $\Rightarrow$ $2|z|>1+z\overline z$

Let $z=x+iy$ then

$$2\sqrt{x^2+y^2}>1+(x+iy)(x-iy)$$

$$2\sqrt{x^2+y^2}>1+(x^2+y^2)$$

$$0>1-2\sqrt{x^2+y^2}+(x^2+y^2)$$

$$0>(1-\sqrt{x^2+y^2})^2$$

since $x,y$ are real number , so there is no solution for this inequality?

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  • $\begingroup$ If you write $z = a + bi$, then you can note that your condition is equivalent to $(a^2 + (b-1)^2 - 2)(a^2 + (b+1)^2 - 2) < 0$, and is two circles with their intersection removed. I don't know of any way to see this easily, but this comment might help someone else. $\endgroup$ – George V. Williams Jan 30 '15 at 22:28
  • $\begingroup$ It should be $\cos 2\theta$ and not $\cos^2\theta$. Comparing the $2$ expressions you get $\frac{1}{r^2} + r^2 < 6-4\cos^2\theta$. How explicit do you need to be? I am not sure you can describe it in words... $\endgroup$ – benji Jan 31 '15 at 4:06
  • $\begingroup$ I always have problem with locus problem like this. I tried to explain and justify every step by algebra and calculating but keep getting into dead end. $\endgroup$ – Diane Vanderwaif Jan 31 '15 at 4:25
  • $\begingroup$ @DianeVanderwaif there must be solutions to the inequality, for example $z=i$ $\endgroup$ – benji Feb 4 '15 at 16:59
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    $\begingroup$ You can use $\Rightarrow$ $\Rightarrow$ or $\implies$ $\implies$. (Whichever you like better.) Of course, you can edit it back to =>, if you prefer it that way. $\endgroup$ – Martin Sleziak Feb 5 '15 at 19:19
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Let $z=re^{i\theta}$. Then after squaring both sides we have $$ 4r^2 > r^4 + 2r^2\cos(2\theta) + 1\Rightarrow 0>(r^2-2r\sin(\theta)-1)(r^2+2r\sin(\theta)-1)\tag{1} $$ where I used the following identity $\cos(2\theta) = 1 -2\sin^2(\theta)$. Let's write $(1)$ in Cartesian coordinates so recall that $r^2 = x^2 + y^2$ and $r\sin(\theta) = y$. Then we have $$ 0>(x^2+y^2-2y-1)(x^2+y^2+2y-1)=[x^2+(y-1)^2-2][x^2+(y+1)^2-2] $$ Then \begin{align} x^2+(y-1)^2&< 2\\ x^2+(y+1)^2&< 2 \end{align} That is, we have two disc of radius $\sqrt{2}$ centered at $(0,\pm i)$. So the solution set is $$ S=\bigl\{z\in\mathbb{C}\colon \lvert z-i\rvert < \sqrt{2}\text{ or }\lvert z+i\rvert < \sqrt{2}\text{ but not in the intersection}\bigr\} $$

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    $\begingroup$ Nice answer. In the final equation, there should be $\sqrt{2}$ and not $2$. $\endgroup$ – mickep Feb 5 '15 at 7:37
  • $\begingroup$ @mickep yes you are correct. I will fix it. $\endgroup$ – dustin Feb 5 '15 at 7:37
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let $D = \{z \colon 2|z| < |1 + z^2|\}.$ observe that the region $D$ is symmetric with respect to the transformations $z \to -z, z \to \bar z$ and $z \to \dfrac{1}{z}$ therefore it is enough to worry about the portion of $D$ in the first quadrant.

Let $o = 0, a = 1$ and $z = re^{it}, p = 1 + r^2e^{2it}$ so that $z$ is on the boundary of $D.$ applying rule of cosine to the triangle, we have $$\cos 2t = 2 - \frac{1}{2}\left( r^2 + \frac{1}{r^2} \right).$$

looking at the graph of $y = 2 - \frac{1}{2}(x + 1/x),$ you can verify that $-1 \le y \le 1$ for $3 - 2\sqrt 2 \le x \le 3 + 2\sqrt 2$ and $f(1) = 1, f(2 \pm \sqrt 3) = 0$

i plotted the region $D$ it contains the unit circle and has a hole(lens) containing the origin.

so part of $D$ in the first quadrant is given by $$\{re^{it} \colon t = \frac{1}{2}\cos^{-1}\left( 2 - \frac{1}{2}( r^2 + \frac{1}{r^2} ) \right), \sqrt{3 - 2\sqrt 2} \le r \le \sqrt{3 + 2 \sqrt 2} \}.$$

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  • $\begingroup$ so my answer is wrong, I wonder where it's wrong? $\endgroup$ – Diane Vanderwaif Feb 2 '15 at 16:43
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    $\begingroup$ @DianeVanderwaif, you have $|1 + z^2 | = 1 +|z|^2$ which is not true except for $z = 1.$ $\endgroup$ – abel Feb 2 '15 at 17:03
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    $\begingroup$ I see, I applied the triangle inequality incorrectly. Thanks $\endgroup$ – Diane Vanderwaif Feb 2 '15 at 17:19
  • $\begingroup$ I'm a little confused . So you let $y=\cos (2t)$ then you use $f(1)$ with $1=r$? Then you let $r^2 \in [3-2\sqrt{2}, 3+2\sqrt{2}]$, when you take the square root,how do you know the sign won't change? because $r$ is always positive? $\endgroup$ – Diane Vanderwaif Feb 2 '15 at 23:31
  • $\begingroup$ and I think you got the inequality sign backward :D $\endgroup$ – Diane Vanderwaif Feb 2 '15 at 23:55

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