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Let “Z” be a complex number then the locus represented by |Z-1| + |Z+2| = 2 is? How could we show effectively that no complex number satisfies this?

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up vote 2 down vote accepted

Here, the most important thing is to understand meaning of modulus value. $|z-a|$ is the distance between $z$ and $a \in \mathbb C$. Now imaine the argand plane. You are asked to find $z$ such that sum of distances of $z$ from $1$ and $-2$ is equal to $2$. Now the minimum sum of distance is when $z$ lies on the line joining these two complex numbers(by triangle law). But distance between these two numbers is $3$, hence the value of LHS can't be less than $3$. e are done.

It would help if you plot these points on argand plane. $1$ is $(1,0$ and so on.

You might want to try something more general. Try to find locus for general constant $c \in \mathbb R^+$.

Now if $c=3$ what do you get?? What if $c > 3$??

A straight line segment in first case. An ellipse in second.

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Forgive me for my poor English first~ Let's convert the problem in a more visually way. In a two-dimensional model, real number on the horizontal axis and imaginary number on the vertical.

$|Z-1| + |Z+2| = 2$ means find a point ($Z$) in this plane while the sum distance of $Z$ to $(1,0)$ and $Z$ to $(-2,0)$ is $2$.

Obviously, the point $Z$ is not exist.

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TheJoker's answer in analytical form: $$|3| = |2 + z - z + 1| \le |2 + z| + |1 - z| = |2|$$ by the triangle inequality. Contradiction.

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Yes, that was what I meant.. Just wanted him to do this work . :-P – TheJoker Nov 15 '12 at 8:47

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