If |z+i|-|z-1|=|z|-2=0 for a complex number z | vedclass

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(B) Given $|z|-2=0 \implies |z|=2$. This represents a circle centered at the origin with radius $2$, so $x^2+y^2=4$. Also, $|z+i|-|z-1|=0 \implies |z+

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$\sqrt{2}(1-i)$

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(B) Given $|z|-2=0 \implies |z|=2$. This represents a circle centered at the origin with radius $2$, so $x^2+y^2=4$. Also, $|z+i|-|z-1|=0 \implies |z+i|=|z-1|$. Let $z=x+iy$. Then $|x+i(y+1)|=|x-1+iy|$. Squaring both sides: $x^2+(y+1)^2=(x-1)^2+y^2$. $x^2+y^2+2y+1=x^2-2x+1+y^2$. $2y=-2x \implies y=-x$. Substituting $y=-x$ into $x^2+y^2=4$: $x^2+(-x)^2=4 \implies 2x^2=4 \implies x^2=2 \implies x=\pm\sqrt{2}$. If $x=\sqrt{2}$, then $y=-\sqrt{2}$, so $z=\sqrt{2}-i\sqrt{2}=\sqrt{2}(1-i)$. If $x=-\sqrt{2}$, then $y=\sqrt{2}$, so $z=-\sqrt{2}+i\sqrt{2}=\sqrt{2}(-1+i)$. Thus, the possible values for $z$ are $\sqrt{2}(1-i)$ and $\sqrt{2}(-1+i)$.

If $|z+i|-|z-1|=|z|-2=0$ for a complex number $z$, then $z=$

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