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(C) Let $z = x + iy$.The given equation is $|z| + |z - 1| = 3$.This represents the sum of distances of a point $z$ from the points $0$ and $1$ being c

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an ellipse

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(C) Let $z = x + iy$.The given equation is $|z| + |z - 1| = 3$.This represents the sum of distances of a point $z$ from the points $0$ and $1$ being constant $(3)$.Since $3 > |0 - 1| = 1$,the locus is an ellipse with foci at $0$ and $1$.Algebraically:$\sqrt{x^2 + y^2} + \sqrt{(x - 1)^2 + y^2} = 3$$\sqrt{(x - 1)^2 + y^2} = 3 - \sqrt{x^2 + y^2}$Squaring both sides:$(x - 1)^2 + y^2 = 9 + x^2 + y^2 - 6\sqrt{x^2 + y^2}$$x^2 - 2x + 1 + y^2 = 9 + x^2 + y^2 - 6\sqrt{x^2 + y^2}$$-2x + 1 = 9 - 6\sqrt{x^2 + y^2}$$6\sqrt{x^2 + y^2} = 2x + 8$$3\sqrt{x^2 + y^2} = x + 4$Squaring again:$9(x^2 + y^2) = (x + 4)^2$$9x^2 + 9y^2 = x^2 + 8x + 16$$8x^2 - 8x + 9y^2 = 16$$8(x^2 - x + \frac{1}{4}) + 9y^2 = 16 + 2$$8(x - \frac{1}{2})^2 + 9y^2 = 18$$\frac{(x - 1/2)^2}{9/4} + \frac{y^2}{2} = 1$This is the standard equation of an ellipse.

The locus of $z$ satisfying $|z|+|z-1|=3$ is

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