Question 233 of 455MediumMCQ
If z is a complex number such that z−1z−i is purely imaginary, then the minimum value of ∣z−(3+3i)∣ is Solutions · 1
Official solution
\[{{z - i} \over {z - 1}}\] is purely imaginary number
Let z=x+iy
\[\therefore\] \[{{x + i(y - 1)} \over {(x - 1) + i(y)}} \times {{(x - 1) - iy} \over {(x - 1) - iy}}\]
\[ \Rightarrow {{x(x - 1) + y(y - 1) + i( - y - x + 1)} \over {{{(x - 1)}^2} + {y^2}}}\] is purely imaginary number
\[ \Rightarrow x(x - 1) + y(y - 1) = 0\]
\[ \Rightarrow {\left( {x - {1 \over 2}} \right)^2} + {\left( {y - {1 \over 2}} \right)^2} = {1 \over 2}\]
\[\therefore\] \[{\left| {z - (3 + 3i)} \right|_{\min }} = \left| {PC} \right| - {1 \over {\sqrt 2 }}\]
\[ = {5 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 2\sqrt 2 \] Answer the question to unlock