The complex number z satisfying the equation | vedclass

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(C) We have,$|z-i|=|z+1|=1$. Let $z=x+iy$. Then $|z-i|=1$ $\Rightarrow |x+i(y-1)|=1$ $\Rightarrow x^2+(y-1)^2=1$ $(i)$. Also,$|z+1|=1$ $\Rightarrow |(

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$-1+i$

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(C) We have,$|z-i|=|z+1|=1$. Let $z=x+iy$. Then $|z-i|=1$ $\Rightarrow |x+i(y-1)|=1$ $\Rightarrow x^2+(y-1)^2=1$ $(i)$. Also,$|z+1|=1$ $\Rightarrow |(x+1)+iy|=1$ $\Rightarrow (x+1)^2+y^2=1$ (ii). Expanding $(i)$: $x^2+y^2-2y+1=1 \Rightarrow x^2+y^2=2y$. Expanding (ii): $x^2+2x+1+y^2=1 \Rightarrow x^2+y^2=-2x$. Equating the two: $2y=-2x \Rightarrow y=-x$. Substituting $y=-x$ into $(i)$: $x^2+(-x-1)^2=1$ $\Rightarrow x^2+x^2+2x+1=1$ $\Rightarrow 2x^2+2x=0$ $\Rightarrow 2x(x+1)=0$. Thus,$x=0$ or $x=-1$. If $x=0$,then $y=0$,so $z=0$. If $x=-1$,then $y=1$,so $z=-1+i$. Therefore,the complex numbers are $0$ and $-1+i$.

The complex number $z$ satisfying the equation $|z-i|=|z+1|=1$ is

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