If 0≤arg⁡(z)≤π/4, then the least value of 2|z−i| is

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If $0 \leq \arg (z) \leq π / 4$ , then the least value of $\sqrt{2} | z - i |$ is

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Detailed Solution

0 $\leq$ arg(z) $\leq$ $π$ / 4 represents the region of the complex plane lying in the Q₁ bounded by the x-axis and the line y = x.
The least value of |z- i| is the length of perpendicular from (0, 1) to y = x, which is $\frac{1}{\sqrt{2}}$ .
Thus, least value of $\sqrt{2} | z - i | = 1$