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Complex Numbers

Section: Mathematics

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Question : 45 of 50

Marks: +1, -0

If a complex number

z

\frac{z-2i}{z-2}

is purely imaginary number and the locus of

z

is a closed curve, then the area of the region bounded by that closed curve and lying in the first quadrant is

[AP EAPCET 19 May 2024 Shift 2]

$2\pi$
$\frac{\pi}{2}$
$\pi$
$\frac{\pi}{4}$

Solution:

\frac{z-2i}{z-2}

is purely imaginary

Let

z=x+iy

\therefore \frac{x+iy-2i}{x+iy-2} = \frac{x+i(y-2)}{(x-2)+iy} \times \frac{(x-2)-iy}{(x-2)-iy}

Real part of

\frac{z-2i}{z-2}=0

x(x-2)+y(y-2)=0

\Rightarrow \; x^2+y^2-2x-2y=0

\Rightarrow (x-1)^2+(y-1)^2=2

, which is equation of circle with radius

\sqrt{2}

units.

Required area

=\frac{1}{2} \times 2 \times 2 + \frac{\pi}{2} (\sqrt{2})^2 = (2+\pi)

sq units

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