What domain of the z-plane is represented by $$|z+2|+|z-2|\lt 4$$ Please give me a hint... Can I use the formula ? $$|z_1|+|z_2|\geqslant |z_1+z_2|$$
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2Think about what this means in words: the set of $z$ such that the sum of the distances from $-2$ and from $2$ are at most $4$. What sort of shape does this remind you of? – PrincessEev Feb 25 '20 at 02:41
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I am confused....is it ellipse? – Ankita Pal Feb 25 '20 at 02:43
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3Hint: think triangle inequality on $|2 + z|$ and $|2 - z|$. – user744868 Feb 25 '20 at 02:43
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1What is the distance from $-2$ to $2$? – J. W. Tanner Feb 25 '20 at 02:48
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@J.W.Tanner, well, that doesn't leave many points to consider $\dots$ – mjw Feb 25 '20 at 02:58
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Take $z_1=2+z$ and $z_2=2-z$ – J. W. Tanner Feb 25 '20 at 03:00
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Note that the LHS of the given equation $$|z+2|+|z-2|\lt 4$$ represents the sum of the distances of the point $z$ to the two fixed points at $(-2,0)$ and $(2,0)$, which has to be less than 4 according to the RHS.
Given that the distance between the two fixed points $(-2,0)$ and $(2,0)$ is already 4, which is the minimum of the LHS, i.e
$$|z+2|+|z-2|\ge |(z+2)-(z-2)| = 4$$
Thus, the given equation represents an empty domain.
Quanto
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Suppose $z_1,z_2\in\mathbb C$ are complex numbers and let $|z|$ denote the modulus of $z$. Then, the triangle inequality states that
$$|z_1+z_2|\le |z_1| + |z_2|$$
similarly, we have
$$|z_1-z_2|=|z_1+(-z_2)|\le|z_1| + |z_2|$$
so let $z_1=z+2, z_2=z-2$ and observe that
$$4=|z_1-z_2|=|(z+2)-(z-2)|\le |z+2| + |z-2|=|z_1| + |z_2|$$
therefore the domain is the empty set.
Axion004
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I understood your method...but can please explain the last line that why this became an empty set? – Ankita Pal Feb 25 '20 at 04:36
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@AnkitaPal - This answer derived the inequality $|z+2|+|z-2|\geq4$, which contradicts your $|z+2|+|z-2|<4$, so there is no $z$ satisfying your inequality. – mr_e_man Feb 25 '20 at 05:07
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