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Complex Numbers Part 1

Section: Mathematics

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Question : 8 of 100

Marks: +1, -0

If z is a complex number such that

\frac{z-i}{z-1}

is purely imaginary, then the minimum value of

|z-(3+3i)|

[31 Aug 2021 Shift 2]

$2\sqrt{2}-1$
$3\sqrt{2}$
$6\sqrt{2}$
$2\sqrt{2}$

Solution:

Let

z=x+iy

\frac{z-i}{z-1}=\frac{x+i(y-1)}{(x-1)+iy}\times\frac{(x-1)-iy}{(x-1)-iy}

=\frac{x(x-1)+y(y-1)}{(x-1)^2+y^2}+i\left[\frac{(x-1)(y-1)-xy}{(x-1)^2+y^2}\right]

As

\frac{z-i}{z-1}

is purely imaginary,

x^2+y^2-x-y=0

\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0

This is a circle with centre

\left(\frac{1}{2},\frac{1}{2}\right)

,

radius

=\frac{1}{2}

which passes through origin as shown in the figure.

Minimum

|z-(3+3i)|=OP-OA

\sqrt{(3-0)^2+(3-0)^2}-\sqrt{\left(\frac{1}{2}-0\right)^2+\left(\frac{1}{2}-0\right)^2}=3\sqrt{2}-\sqrt{2}=2\sqrt{2}

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