If the complex number z=x+iy,where i=sqrt{-1} | vedclass

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(C) Given the condition $|z+1|=1$. Substitute $z=x+iy$ into the equation: $|x+iy+1|=1$ $|(x+1)+iy|=1$ Using the definition of the modulus of a complex

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circle with centre $(-1,0)$ and radius $1$ unit.

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(C) Given the condition $|z+1|=1$. Substitute $z=x+iy$ into the equation: $|x+iy+1|=1$ $|(x+1)+iy|=1$ Using the definition of the modulus of a complex number $|a+ib|=\sqrt{a^2+b^2}$,we get: $\sqrt{(x+1)^2+y^2}=1$ Squaring both sides: $(x+1)^2+y^2=1^2$ $(x+1)^2+y^2=1$ This is the standard equation of a circle $(x-h)^2+(y-k)^2=r^2$,where the centre is $(h,k)$ and the radius is $r$. Comparing the equations,the centre is $(-1,0)$ and the radius is $1$ unit.

If the complex number $z=x+iy$,where $i=\sqrt{-1}$,satisfies the condition $|z+1|=1$,then $z$ lies on

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