The complex number $z$ satisfying the equations $\left| z \right| -4=\left| z-i \right| -\left| z+5i \right| =0$, is

We have two equation $\left| z \right| -4=0$ and $\left| z-i \right| -\left| z+5i \right| =0$

Putting $z=x+iy$ i.e ${ x }^{ 2 }+{ y }^{ 2 }=16$ ....(1)

and $\left| x+iy-i \right| =\left| x+iy+5i \right|$

$\Rightarrow{ x }^{ 2 }+{ \left( y-1 \right) }^{ 2 }={ x }^{ 2 }+{ \left( y+5 \right) }^{ 2 }$

$\Rightarrow y=-2$ ...(2)

Putting $y=-2$ in (1),

${ x }^{ 2 }+4=16\Rightarrow x=\pm 2\sqrt { 3 }$

Hence the complex numbers $z$ satisfying the given equation are

${ z }_{ 1 }=2\sqrt { 3 } -2i$ and ${ z }_{ 2 }=-2\sqrt { 3 } -2i$