Match the statements in Column I with those in Column II.
[Note : Here z takes values in the complex

Q: Why do we use Euler's form instead of just $z = x + iy$ for everything?

Using $z = x + iy$ often leads to messy algebraic nightmares, especially with complex transformations. Euler's form $w = re^{i\theta}$ is the 'skeleton key' of complex numbers. It transforms complex constraints into simple trigonometric identities, making the locus obvious and elegant.

Q: In the first part, how did we know the locus was the Real Axis?

The equation $|z - i|z|| = |z + i|z||$ implies $z$ is equidistant from $i|z|$ and $-i|z|$. Since these two points are symmetric about the origin on the imaginary axis, the set of points equidistant from them must be the perpendicular bisector—which is the Real Axis!

Q: For $z = w + 1/w$ with $|w|=1$, why is the locus a line segment and not a full ellipse?

Great observation! When $|w|=1$, $w = e^{i\theta}$. Then $z = e^{i\theta} + e^{-i\theta} = 2\cos\theta$. Since $\cos\theta$ is bounded between $-1$ and $1$, $z$ is restricted to the interval $[-2, 2]$ on the real axis. It is a degenerate ellipse, effectively a line segment.

Question

Match the statements in Column I with those in Column II.
[Note : Here $z$ takes values in the complex plane and $Im z$ and $Re z$ denote , respectively, the imaginary part and the real part of $z$. ]

List-I:

P. The set of points $z$ satisfying $|z - i|z|| = |z + i|z||$ is contained in or equal to

Q. The set of points $z$ satisfying $|z + 4| + |z - 4| = 10$ is contained in or equal to

R. If $|w| = 2$, then the set of points $z = w - \frac{1}{w}$ is contained in or equal to

S. If $|w| = 1$, then the set of points $z = w + \frac{1}{w}$ is contained in or equal to

List-II:

1. an ellipse with eccentricity $\frac{4}{5}$

2. the set of points $z$ satisfying $Im z = 0$

3. the set of points $z$ satisfying $|Im z| \le 1$

4. the set of points $z$ satisfying $|Re z| < 2$

5. the set of points $z$ satisfying $|z| \le 3$

Sigma Percentile · Accepted Answer

### The Perpendicular Bisector: A Geometric Insight

We begin with the equation $|z - i|z|| = |z + i|z||$. The moment you see an equation of the form $|z - z_1| = |z - z_2|$, your brain should immediately recognize the **Perpendicular Bisector**.

This is the locus of all points $z$ that are equidistant from two fixed points, $z_1$ and $z_2$. In our case, $z_1 = i|z|$ and $z_2 = -i|z|$.

Visualize the complex plane. These points lie on the imaginary axis, perfectly symmetric above and below the origin. If you are standing anywhere on the Real Axis, you are equidistant from these two points.

Thus, the locus is the Real Axis, where $Im(z) = 0$. This immediately satisfies the condition $Im(z) = 0$ and, by extension, $|Im(z)| \le 1$.

### The Ellipse: The Dance of Foci

Next, we encounter the equation:
$$|z + 4| + |z - 4| = 10$$

This is a classic. The definition of an ellipse is the locus of points where the sum of distances from two fixed points (the foci) is constant. Here, the foci are at $-4$ and $4$, and the constant sum is $2a = 10$.

This gives us a semi-major axis $a = 5$. The distance between the foci is $2c = 8$, so $c = 4$. The eccentricity $e$ is defined as:
$$e = \frac{c}{a} = \frac{4}{5}$$

### The Transformation: $z = w - 1/w$

Now, things get interesting. We are given $|w| = 2$ and a transformation $z = w - 1/w$. Whenever you see a modulus constraint like $|w| = 2$, reach for Euler's form: $w = 2e^{i	heta}$.

Substituting this, we get:
$$z = 2e^{i	heta} - \frac{1}{2}e^{-i	heta}$$

Using Euler's identity $e^{i	heta} = \cos	heta + i\sin	heta$, we expand this to:
$$z = 2(\cos	heta + i\sin	heta) - \frac{1}{2}(\cos	heta - i\sin	heta)$$

Grouping the real and imaginary parts, we find:
$$Re(z) = \frac{3}{2}\cos	heta, \quad Im(z) = \frac{5}{2}\sin	heta$$

If we set $x = \frac{3}{2}\cos	heta$ and $y = \frac{5}{2}\sin	heta$, we can eliminate $	heta$ using $\cos^2	heta + \sin^2	heta = 1$. This yields the equation of an ellipse:
$$\frac{x^2}{(3/2)^2} + \frac{y^2}{(5/2)^2} = 1$$

This ellipse has an eccentricity of $4/5$, its real part is bounded by $1.5$, and its total modulus is bounded by $2.5$.

### The Degenerate Case: $z = w + 1/w$

Finally, we look at $z = w + 1/w$ where $|w| = 1$. Again, use $w = e^{i	heta}$. Then:
$$z = e^{i	heta} + e^{-i	heta}$$

This simplifies beautifully to $z = 2\cos	heta$. Notice that the imaginary part is zero.

The entire locus is just a line segment on the real axis from $-2$ to $2$. Since it lies on the real axis, $Im(z) = 0$, and since it is bounded between $-2$ and $2$, it satisfies $|Re(z)| < 2$ and $|z| \le 3$.

Animated Solution for Mathematics - Complex Numbers: Match the statements in Column I with those in Column II. [Note : Here $z$ takes values in the complex plane and $I m z$ and $R ez$ denote , respectively, the imaginary part and the real part of $z$ . ]

List-I

List-II

Select Matching Pairs:

Visualized Solution

Analyzing the Complex Loci

Column I(A): $∣ z - i ∣ z ∣∣ = ∣ z + i ∣ z ∣∣$

Geometric Interpretation of (A)

Matching Column I(A)

Column I(B): $∣ z + 4∣ + ∣ z - 4∣ = 10$

Eccentricity of Ellipse (B)

Column I(C): $z = w - \frac{1}{w}$

Expanding into Real and Imaginary Parts

Locus of (C)

Matching Column I(C)

Column I(D): $z = w + \frac{1}{w}$

Simplifying Locus (D)

Matching Column I(D)

The Sigma Insight: Geometrical Applications of Complex Numbers

The Perpendicular Bisector

The Ellipse

The Transformation: $z = w - 1/ w$

The Degenerate Case: $z = w + 1/ w$

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List-I

List-II

Select Matching Pairs:

Visualized Solution

Analyzing the Complex Loci

Column I(A): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣z−i∣z∣∣=∣z+i∣z∣∣

Geometric Interpretation of (A)

Matching Column I(A)

Column I(B): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣z+4∣+∣z−4∣=10

Eccentricity of Ellipse (B)

Column I(C): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=w−w1​

Expanding into Real and Imaginary Parts

Locus of (C)

Matching Column I(C)

Column I(D): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=w+w1​

Simplifying Locus (D)

Matching Column I(D)

The Sigma Insight: Geometrical Applications of Complex Numbers

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The Ellipse

The Transformation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=w−1/w

The Degenerate Case: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=w+1/w

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