If `omega=z/(z-1/3i)` and `abs(omega)=1`, where `i=sqrt(-1)`,then lies on

If z is a complex number which simultaneously satisfies the equations 3abs(z-12)=5abs(z-8i) " and " abs(z-4) =abs(z-8) , where i=sqrt(-1) , then Im(z) can be

If omega = z//[z-(1//3)i] and |omega| = 1 , then find the locus of z.

If (z+1)/(z+i) is a purely imaginary number (where (i=sqrt(-1) ), then z lies on a

If z=(i) ^((i) ^(i)) where i= sqrt (−1) ​ , then z is equal to

If z is any complex number satisfying abs(z-3-2i) le 2 , where i=sqrt(-1) , then the maximum value of abs(2z-6+5i) , is

The equation z^(2)-i|z-1|^(2)=0, where i=sqrt(-1), has.

If w= z/(z-i1/3) and |w|=1,then z lies on:

If |z-2i|lesqrt(2), where i=sqrt(-1), then the maximum value of |3-i(z-1)|, is

If (3+i)(z+bar(z))-(2+i)(z-bar(z))+14i=0 , where i=sqrt(-1) , then z bar(z) is equal to

If z=(3+4i)^(6)+(3-4i)^(6),"where" i=sqrt(-1), then Find the value of Im(z) .