Consider the region S of complex numbers a such that `|z^(2) - az + 1|=1`, where `|z|=1` . Then area of S in the Argand plane is

Options

A. `pi+8`
B. `pi + 4`
C. `2pi+ 4`
D. `pi + 6`

Answer

A

Explanation

Given `|z|= 1`
and `|z^(2) - az + 1|=1`
`rArr |z||z-a +(1)/(2)|= 1`
`rArr = |a-(z+(1)/(2))|= 1`
Now `z = cos theta + sin theta`
`rArr z + (1)/(2)= 2 cos theta`
So, (1) readuces to
`|a-2cos theta|=1`
So, loucs of a is circle whose centre is `(2 cos theta,0)` and radius is 1.
Since, `-2 le 2 cos theta le 2`, we have following region traced by the locus of a.

As shown in the figure the set S consists of the locus of units circles centred at a point on the line segment with end points at -2 and 2 on the Argnd plane. Therefore, the area S consists of the area of two semi circles and area of rectangle ABCD.
So, the required area is `pi + 8` .