Question

If $S = \left\{z \in \mathbb{C} : \frac{z - i}{z + 2i} \in \mathbb{R}\right\}$, then :

• A. S contains only one element
• B. S contains exactly two elements
• C. S is a straight line in the complex plane
• D. S is a circle in the complex plane

Hint:

If $\frac{z-z_1}{z-z_2}$ is real, $z$ lies on the line passing through $z_1$ and $z_2$. Here $z_1 = i$ and $z_2 = -2i$, which are both on the imaginary axis.

Answer 1

The Correct Option is C

To solve the problem, we need to determine the nature of the set \( S \) defined by:

S = \left\{ z \in \mathbb{C} : \frac{z - i}{z + 2i} \in \mathbb{R} \right\}

This expression states that for a complex number \( z \), the fraction \frac{z - i}{z + 2i} must be a real number. A complex number is real if its imaginary part is zero. Hence, we need to find all \( z = x + yi \) (where \( x, y \in \mathbb{R} \)) such that the imaginary part of \(\frac{z - i}{z + 2i}\) is zero.

Substitute \( z = x + yi \):

\frac{z - i}{z + 2i} = \frac{(x + yi) - i}{(x + yi) + 2i}

This simplifies to:

\frac{(x + y - 1)i}{x + (y + 2)i}

To find when this fraction is real, we consider its conjugate form, which gives a real number when the imaginary part of \( (x - 1 - yi)(x + (y + 2)i)^* \) is zero. Simplifying gives:

\text{Imaginary part} = -5x + x^2 + 2 + (y + 2)^2 - 1 - yx = 0

The imaginary part is zero if:

y(x + 2) = -2(x - 1)

Solving this linear equation, substitute and simplify:

yx + 2y = -2x + 2

yx + 2y + 2x - 2 = 0

x(y + 2) = 2(y + 1)

x = \frac{2(y + 1)}{y + 2}

This is the equation of a straight line in the complex plane. Hence, the set \( S \) corresponds to a straight line in the complex plane. Therefore, the correct answer is:

S is a straight line in the complex plane.