Q48 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Mathematics

For any complex number $z$, the minimum value of $|z|+|z-1|$ is

  • Correct answer
    A
    1
  • B
    0
  • C
    $1 / 2$
  • D
    $3 / 2$

Solution

Let $z=x+i y$ $$ \begin{aligned} |z| & =\sqrt{x^2+y^2} \\ |z-1| & =\sqrt{(x-1)^2+y^2}=\sqrt{x^2-2 x+1+y^2} \end{aligned} $$ For minimum value of $|z|$ and $|z-1|$ $$ \begin{aligned} & |z-1| \geq 0 \\ & x-1 \geq 0 \left[\because y^2 \geq 0\right]\\ & x \geq 1 \end{aligned} $$ The minimum value of $|z|+|z-1|$ is 1 .

Hence, the correct answer is option: 'A'.

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