Question #2831

Find the locus of points in the plane satisfying each of the given conditions:
<br>(i) |z-5| = 6 <br> (ii) |z-2i| >=1 <br> (iii) Re(z+2) = -1
<br>(iv) Re{i(conjugate of z)} =3 <br>(v) |z+i| = |z-i|
<br>Also Sketch its diagram.
Please tell me as soon as possible...?

Expert's answer

(i) |z-5| = 6

This is a circle of radius 6 with center at 5

(ii) |z-2i| >=1

This is a outer part of a circle of radius 1 with center at 2i together with this circle

(iii) Re(z+2) = -1

This is a vertical line x+2=-1, that is x=3.

(iv) Re{i(conjugate of z)} =3

Let z=x+iy, then

i(conjugate of z) = ix-y,

whence

3 = Re{i(conjugate of z)} =

= Re{-y+ix}= -y

Thus we obtain a horizontal line y=-3.

(v) |z+i| = |z-i|

or any r the set

& |z+i|=r and |z-i|=r

Are two circles of radius r centered at -i and I respectively.

Then |z+i| = |z-i| is the set of all points equidistant to i and –i

This is a real axis y=0.

This is a circle of radius 6 with center at 5

(ii) |z-2i| >=1

This is a outer part of a circle of radius 1 with center at 2i together with this circle

(iii) Re(z+2) = -1

This is a vertical line x+2=-1, that is x=3.

(iv) Re{i(conjugate of z)} =3

Let z=x+iy, then

i(conjugate of z) = ix-y,

whence

3 = Re{i(conjugate of z)} =

= Re{-y+ix}= -y

Thus we obtain a horizontal line y=-3.

(v) |z+i| = |z-i|

or any r the set

& |z+i|=r and |z-i|=r

Are two circles of radius r centered at -i and I respectively.

Then |z+i| = |z-i| is the set of all points equidistant to i and –i

This is a real axis y=0.

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