If z satisfies the inequality `|z-1-2i|<=1`, then

The locus of z satisfying the inequality |(z+2i)/(2z+i)|lt1 where z = x + iy , is :

The locus of z satisfying the inequality |(z+2i)/(2z+1)|<1, where z=x+iy, is :

On the complex plane locus of a point z satisfying the inequality 2<=|z-1|<3 denotes

If |z-3i| ltsqrt5 then prove that the complex number z also satisfies the inequality |i(z+1)+1| lt2sqrt5 .

The locus of z which satisfies the inequality log _(0.3) abs(z-1) gt log _(0.3) abs(z-i) is given by :

The locus of z which satisfies the inequality log _(0.3) abs(z-1) gt log _(0.3) abs(z-i) is given by :

If the point ((k-1)/k,(k-2)/k) lies on the locus of z satisfying the inequality |(z+3i)/(3z+i)| lt 1 , then the interval in which k lies is

The least value of |z| where z is complex number which satisfies the inequality exp (((|z|+3)(|z|-1))/(|z|+1|)log_(e)2)gtlog_(sqrt2)|5sqrt7+9i|,i=sqrt(-1) is equal to :