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For any complex number $z$ , the minimum value of $\left| z \right|+\left| z-1 \right|$ is
  & \left( A \right)0 \\
 & \left( B \right)1 \\
 & \left( C \right)2 \\
 & \left( D \right)-1 \\

Last updated date: 20th Jun 2024
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Hint: Here, we need to apply the triangular inequality after rewriting the given expression as $\left| z \right|+\left| -\left( z-1 \right) \right|$ . The inequality of will then be $\left| z \right|+\left| -\left( z-1 \right) \right|\ge \left| z+\left( -\left( z-1 \right) \right) \right|$ which has the minimum value $1$ .

Complete step by step answer:
Any complex number $z$ can be represented as $x+yi$ where, $x$ is the real part and $y$ is the imaginary part. $\left| z \right|$ is represented as $\sqrt{{{x}^{2}}+{{y}^{2}}}$ and gives the intuition of the distant of a point $\left( x,y \right)$ from the origin. Similarly, $\left| z-1 \right|$ is represented as $\sqrt{{{\left( x-1 \right)}^{2}}+{{y}^{2}}}$ and gives the intuition of the distance of the point $\left( x,y \right)$ from the point $\left( 1,0 \right)$ .
The given expression is $\left| z \right|+\left| z-1 \right|$ . We need to minimise this expression. In order to minimize this expression, we can take the help of the triangle inequality which states that
$\left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|\ge \left| {{z}_{1}}+{{z}_{2}} \right|$
This inequality can be understood by a simple logic. The most general case would be ${{z}_{1}}$ being completely positive and ${{z}_{2}}$ being negative. Taking their absolute values separately and then adding the two would mean simply adding two positive numbers. But, if we first add them and then take their absolute values, we will always get a smaller number as the negative ${{z}_{2}}$ will cancel some part of the positive ${{z}_{1}}$and their result will be smaller.
To apply the triangle inequality in the given problem, we first need to rewrite the expression as
 $\Rightarrow \left| z \right|+\left| z-1 \right|=\left| z \right|+\left| -\left( z-1 \right) \right|$
We now apply the triangle inequality as,
$\Rightarrow \left| z \right|+\left| -\left( z-1 \right) \right|\ge \left| z+\left( -\left( z-1 \right) \right) \right|$
Simplifying the above expression, we get,
$\Rightarrow \left| z \right|+\left| -\left( z-1 \right) \right|\ge \left| z+\left( 1-z \right) \right|$
Opening up the brackets, and then carrying out the subtraction, we get,
  & \Rightarrow \left| z \right|+\left| -\left( z-1 \right) \right|\ge \left| z+1-z \right| \\
 & \Rightarrow \left| z \right|+\left| -\left( z-1 \right) \right|\ge \left| 1 \right| \\
We know that $\left| 1 \right|$ is nothing but $1$ . The inequality thus becomes,
$\left| z \right|+\left| z-1 \right|\ge 1$
Therefore, we can conclude that the minimum value of the given expression $\left| z \right|+\left| z-1 \right|$ is $1$ , that is option $\left( B \right)$ .

Note: These types of problems are tricky and require correct rewriting of the expression to get the desired answer. For example, if we write the inequality as $\left| z \right|+\left| z-1 \right|\ge \left| z+z-1 \right|$ which becomes $\Rightarrow \left| z \right|+\left| z-1 \right|\ge \left| 2z-1 \right|$ , which has the minimum value $0$ which is not the correct answer. This problem can also be solved in $x,y$ terms, but it will become tedious.