Match Complex Number Statements with Geometric Regions in Complex Plane

To match the statements in Column-I with the appropriate regions or properties in Column-II, we analyze each case individually:

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Statement A:

The set of points $z$ satisfying $|z - i|z|| = |z + i|z||$.

Let $z = x + iy$, where $x, y \in \mathbb{R}$ and $|z| = r = \sqrt{x^2 + y^2}$.
Substituting $z$ into the equation gives:
$|x + i(y - r)| = |x + i(y + r)|$

Squaring both sides:
$x^2 + (y - r)^2 = x^2 + (y + r)^2$
$(y - r)^2 = (y + r)^2$
$y^2 - 2yr + r^2 = y^2 + 2yr + r^2$
$4yr = 0$

This implies $y = 0$ or $r = 0$. In either case, we have:
$y = \text{Im } z = 0$

Thus, the set of points is the real axis ($\text{Im } z = 0$).

Checking Column-II options:

• q) Since $\text{Im } z = 0$, it is equal to the set in q.
• r) Since $\text{Im } z = 0$, the inequality $|\text{Im } z| = 0 \le 1$ holds for all points in the set, so it is contained in r.

Therefore, A matches with q, r.

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Statement B:

The set of points $z$ satisfying $|z + 4| + |z - 4| = 10$.

This equation represents the locus of a point $z$ such that the sum of its distances from two fixed points (foci) $S_1(-4, 0)$ and $S_2(4, 0)$ is a constant equal to $10$.

Since the distance between the foci is $2ae = 8$ and $2a = 10$, we have:
$a = 5, \quad ae = 4 \implies e = \frac{4}{5}$

The semi-minor axis $b$ is given by:
$b^2 = a^2(1 - e^2) = 25\left(1 - \frac{16}{25}\right) = 9 \implies b = 3$

The locus is an ellipse with standard equation $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and eccentricity $e = \frac{4}{5}$.

Checking Column-II options:

• p) The locus is an ellipse with eccentricity $\frac{4}{5}$.

Therefore, B matches with p.

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Statement C:

If $|w| = 2$, then $z = w - \frac{1}{w}$.

Let $w = 2e^{i\theta} = 2(\cos\theta + i\sin\theta)$.
Then:
$\frac{1}{w} = \frac{1}{2}e^{-i\theta} = \frac{1}{2}(\cos\theta - i\sin\theta)$

Substituting into $z$:
$z = \left(2 - \frac{1}{2}\right)\cos\theta + i\left(2 + \frac{1}{2}\right)\sin\theta = \frac{3}{2}\cos\theta + i \frac{5}{2}\sin\theta$

Let $z = x + iy$:
$x = \frac{3}{2}\cos\theta, \quad y = \frac{5}{2}\sin\theta$

Eliminating $\theta$ using $\cos^2\theta + \sin^2\theta = 1$:
$\frac{x^2}{(3/2)^2} + \frac{y^2}{(5/2)^2} = 1$

This is the equation of an ellipse centered at the origin with semi-axes $a = \frac{5}{2}$ (along the imaginary axis) and $b = \frac{3}{2}$ (along the real axis).
The eccentricity $e$ is:
$b^2 = a^2(1 - e^2) \implies \left(\frac{3}{2}\right)^2 = \left(\frac{5}{2}\right)^2 (1 - e^2) \implies \frac{9}{25} = 1 - e^2 \implies e = \frac{4}{5}$

Now checking bounds on $z$:

• p) It is an ellipse with eccentricity $\frac{4}{5}$.
• s) $|\text{Re } z| = \left|\frac{3}{2}\cos\theta\right| \le \frac{3}{2} \le 2$.
• t) $|z| = \sqrt{x^2 + y^2} \le \sqrt{\left(\frac{3}{2}\right)^2 \cos^2\theta + \left(\frac{5}{2}\right)^2 \sin^2\theta} \le \frac{5}{2} \le 3$.

Therefore, C matches with p, s, t.

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Statement D:

If $|w| = 1$, then $z = w + \frac{1}{w}$.

Let $w = e^{i\theta} = \cos\theta + i\sin\theta$.
Then:
$z = e^{i\theta} + e^{-i\theta} = 2\cos\theta$

Since $\cos\theta \in [-1, 1]$, the set of points $z$ is the line segment $[-2, 2]$ on the real axis ($\text{Im } z = 0$).

Checking Column-II options:

• q) Since $z \in \mathbb{R}$, $\text{Im } z = 0$.
• r) $|\text{Im } z| = 0 \le 1$.
• s) $|\text{Re } z| = |2\cos\theta| \le 2$.
• t) $|z| = |2\cos\theta| \le 2 \le 3$.

Therefore, D matches with q, r, s, t.

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Final Match:

• A $\rightarrow$ q, r
• B $\rightarrow$ p
• C $\rightarrow$ p, s, t
• D $\rightarrow$ q, r, s, t