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(C) Given $z = x + iy$. The equation is $|z - 1| + |z + i| = 2$. Substituting $z = x + iy$,we get $|(x - 1) + iy| + |x + i(y + 1)| = 2$. This implies

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$3x^2 - 2xy + 3y^2 - 4x + 4y = 0$

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(C) Given $z = x + iy$. The equation is $|z - 1| + |z + i| = 2$. Substituting $z = x + iy$,we get $|(x - 1) + iy| + |x + i(y + 1)| = 2$. This implies $\sqrt{(x - 1)^2 + y^2} + \sqrt{x^2 + (y + 1)^2} = 2$. Rearranging,$\sqrt{(x - 1)^2 + y^2} = 2 - \sqrt{x^2 + (y + 1)^2}$. Squaring both sides: $(x - 1)^2 + y^2 = 4 + x^2 + (y + 1)^2 - 4\sqrt{x^2 + (y + 1)^2}$. $x^2 - 2x + 1 + y^2 = 4 + x^2 + y^2 + 2y + 1 - 4\sqrt{x^2 + (y + 1)^2}$. $-2x - 2y - 4 = -4\sqrt{x^2 + (y + 1)^2}$. Dividing by $-2$: $x + y + 2 = 2\sqrt{x^2 + (y + 1)^2}$. Squaring both sides again: $(x + y + 2)^2 = 4(x^2 + y^2 + 2y + 1)$. $x^2 + y^2 + 4 + 2xy + 4x + 4y = 4x^2 + 4y^2 + 8y + 4$. Simplifying: $3x^2 - 2xy + 3y^2 - 4x + 4y = 0$.

If $z = x + iy$ and the point $P$ represents $z$ in the Argand plane,then the locus of $z$ satisfying the equation $|z - 1| + |z + i| = 2$ is

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