If z is a complex number in the Argand plane, | vedclass

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(B) The given equation is $|z - 2| + |z + 2| = 8$.This is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = 2$ and $z_2 = -2$.Here,the distance be

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Ellipse

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(B) The given equation is $|z - 2| + |z + 2| = 8$.This is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = 2$ and $z_2 = -2$.Here,the distance between the fixed points (foci) is $|z_1 - z_2| = |2 - (-2)| = 4$.Since the sum of distances from two fixed points is constant $(8)$ and this constant is greater than the distance between the fixed points $(4)$,the locus of $z$ is an ellipse.Algebraically,let $z = x + iy$:$\sqrt{(x - 2)^2 + y^2} + \sqrt{(x + 2)^2 + y^2} = 8$$\sqrt{(x - 2)^2 + y^2} = 8 - \sqrt{(x + 2)^2 + y^2}$Squaring both sides:$(x - 2)^2 + y^2 = 64 + (x + 2)^2 + y^2 - 16\sqrt{(x + 2)^2 + y^2}$$x^2 - 4x + 4 + y^2 = 64 + x^2 + 4x + 4 + y^2 - 16\sqrt{(x + 2)^2 + y^2}$$-8x - 64 = -16\sqrt{(x + 2)^2 + y^2}$$x + 8 = 2\sqrt{(x + 2)^2 + y^2}$Squaring again:$x^2 + 16x + 64 = 4(x^2 + 4x + 4 + y^2)$$x^2 + 16x + 64 = 4x^2 + 16x + 16 + 4y^2$$3x^2 + 4y^2 = 48$$\frac{x^2}{16} + \frac{y^2}{12} = 1$,which is the equation of an ellipse.

If $z$ is a complex number in the Argand plane,then the equation $|z - 2| + |z + 2| = 8$ represents

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