If z = x + iy satisfies |z|-2=0 and |z-i|-|z+ | vedclass

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Question

$|z-i|-|z+5 i|=0$

$|x+(y-1) i|=|x+(y+5) i|$

$x^{2}+(y-1)^{2}=x^{2}+(y+5)^{2}$

$(y-1)^{2}-(y+5)^{2}=0$

$(2 y+4)(-6)=0$

$y=-2$

$	here

Vedclass · Accepted Answer

$x^{2}+y-4=0$

Vedclass · Answer

$|z-i|-|z+5 i|=0$

$|x+(y-1) i|=|x+(y+5) i|$

$x^{2}+(y-1)^{2}=x^{2}+(y+5)^{2}$

$(y-1)^{2}-(y+5)^{2}=0$

$(2 y+4)(-6)=0$

$y=-2$

$	herefore x^{2}+(-2)^{2}=4$

$x=0$

$Z \equiv(0,-2)$, check options

If $z = x + iy$ satisfies $|z|-2=0$ and $|z-i|-|z+5 i|=0$, then

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