The minimum value of $| z | + | z - 1 | + | z - 2 |$ is

0

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4

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Solution

The correct option is C $2$
Let $z = x + i y$
$f (z) = | z | + | z - 1 | + | z - 2 |$
$f (x, y) = \sqrt{x^{2} + y^{2}} + \sqrt{(x - 1)^{2} + y^{2}} + \sqrt{(x - 2)^{2} + y^{2}}$
For minimum value, $y = 0$
$f (x) = | x | + | x - 1 | + | x - 2 |$ for all $x$
The function can be expressed as follows
$f (x) = 3 x - 3$ , $x \geq 2$
$f (x) = x + 1,$ $1 \leq x < 2$
$f (x) = 3 - x$ , $0 \leq x \leq 1$
$f (x) = - 3 x + 3$ , $x < 0$
Hence, $f (m i n) = f (1) = 2$

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Similar questions

| z - 1 | + | z - 2 | + | z - 3 | + | z - 4 | + | z - 5 |

f (z) = | z | + | z - 1 | + | z + 2 |

z_{1} = 5 + 12 i

| z_{2} | = 4

\frac{4 z_{1}}{9 z_{2}} + \frac{4_{1}}{9_{2}} = 0

∣ ∣ ∣ \frac{z_{1} - z_{2}}{z_{1} + z_{2}} ∣ ∣ ∣

z_{1}, z_{2}

| z_{1} | = 12

| z_{2} - 3 - 4 i | = 5

| z_{1} - z_{2} |

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