If S = \ z \in C:z - i \over z + 2i \in R \ , then :

S is a straight line in the complex plane

S contains exactly two elements

S is a circle in the complex plane

Complex Numbers PYQ — JEE Main 2021 · 27 Aug · Shift 1 · Q26 with Solution

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Correct answer: D

We need the set $S=\left\{z\in \mathbb C:\frac{z-i}{z+2i}\in \mathbb R\right\}.$

This means the complex number $\frac{z-i}{z+2i}$ is real.

Let $z=x+iy,$ where $x,y\in \mathbb R$ . Then $z-i=x+i(y-1),$ $z+2i=x+i(y+2).$ So the condition is $\frac{x+i(y-1)}{x+i(y+2)}\in \mathbb R.$
A quotient of two complex numbers is real iff its imaginary part is zero. So rationalize:

=\frac{(x+i(y-1))(x-i(y+2))}{x^2+(y+2)^2}.$$ Now expand the numerator:

(x+i(y-1))(x-i(y+2)) = x^2 -ix(y+2)+ix(y-1)+ (i(y-1))(-i(y+2)).

Since $i\cdot (-i)=1$, $$ (i(y-1))(-i(y+2))=(y-1)(y+2). $$ Also, $$-ix(y+2)+ix(y-1)=ix[(y-1)-(y+2)]=-3ix.$$ Thus numerator becomes $$x^2+(y-1)(y+2)-3ix.$$ Therefore, $$ rac{z-i}{z+2i} =\frac{x^2+(y-1)(y+2)}{x^2+(y+2)^2}-i\frac{3x}{x^2+(y+2)^2}.$$ For this to be real, imaginary part must be zero: $$-\frac{3x}{x^2+(y+2)^2}=0.$$ Hence $$x=0.$$ 4. So all points in $S$ satisfy $$z=iy,$$ which is the imaginary axis. But we must also ensure the denominator is nonzero: $$z+2i\ne 0 \implies z\ne -2i.$$ So the set is the line $x=0$ with the point $-2i$ excluded. 5. Geometric interpretation: The points $i$ and $-2i$ are fixed points on the imaginary axis. The condition $$\frac{z-i}{z+2i}\in \mathbb R$$ means $(z-i)$ and $(z+2i)$ have the same or opposite arguments, so the points $z, i, -2i$ are collinear. Hence $z$ lies on the straight line through $i$ and $-2i$, i.e. the imaginary axis, excluding $z=-2i$. 6. Evaluate options: - A: "exactly two elements" — false. - B: "only one element" — false. - C: "a circle" — false. - D: "a straight line in the complex plane" — true (more precisely, a straight line with one point removed). Therefore, the correct option is $$\boxed{D}.$$

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View written solutionFree

Correct answer: D

We need the set $S=\left\{z\in \mathbb C:\frac{z-i}{z+2i}\in \mathbb R\right\}.$

This means the complex number $\frac{z-i}{z+2i}$ is real.

Let $z=x+iy,$ where $x,y\in \mathbb R$ . Then $z-i=x+i(y-1),$ $z+2i=x+i(y+2).$ So the condition is $\frac{x+i(y-1)}{x+i(y+2)}\in \mathbb R.$
A quotient of two complex numbers is real iff its imaginary part is zero. So rationalize:

=\frac{(x+i(y-1))(x-i(y+2))}{x^2+(y+2)^2}.$$ Now expand the numerator:

(x+i(y-1))(x-i(y+2)) = x^2 -ix(y+2)+ix(y-1)+ (i(y-1))(-i(y+2)).

Since $i\cdot (-i)=1$, $$ (i(y-1))(-i(y+2))=(y-1)(y+2). $$ Also, $$-ix(y+2)+ix(y-1)=ix[(y-1)-(y+2)]=-3ix.$$ Thus numerator becomes $$x^2+(y-1)(y+2)-3ix.$$ Therefore, $$ rac{z-i}{z+2i} =\frac{x^2+(y-1)(y+2)}{x^2+(y+2)^2}-i\frac{3x}{x^2+(y+2)^2}.$$ For this to be real, imaginary part must be zero: $$-\frac{3x}{x^2+(y+2)^2}=0.$$ Hence $$x=0.$$ 4. So all points in $S$ satisfy $$z=iy,$$ which is the imaginary axis. But we must also ensure the denominator is nonzero: $$z+2i\ne 0 \implies z\ne -2i.$$ So the set is the line $x=0$ with the point $-2i$ excluded. 5. Geometric interpretation: The points $i$ and $-2i$ are fixed points on the imaginary axis. The condition $$\frac{z-i}{z+2i}\in \mathbb R$$ means $(z-i)$ and $(z+2i)$ have the same or opposite arguments, so the points $z, i, -2i$ are collinear. Hence $z$ lies on the straight line through $i$ and $-2i$, i.e. the imaginary axis, excluding $z=-2i$. 6. Evaluate options: - A: "exactly two elements" — false. - B: "only one element" — false. - C: "a circle" — false. - D: "a straight line in the complex plane" — true (more precisely, a straight line with one point removed). Therefore, the correct option is $$\boxed{D}.$$