If Re((z-1)/(2z+i)) = 1, where z = x + iy, then the point (x, y) lies on a (A) circle whose centre i

Q: Why do we multiply by the conjugate? Can't we just equate the real parts directly?

That is a classic trap! You cannot simply take the real part of the numerator and divide it by the real part of the denominator. The real part of a complex fraction $\frac{z_1}{z_2}$ is $\frac{Re(z_1 \cdot \bar{z_2})}{|z_2|^2}$. By multiplying by the conjugate, you are essentially 'rationalizing' the denominator, making it a real number so that the real part of the entire fraction becomes clearly visible in the numerator.

Q: Wait, why did the $i$ disappear in the numerator when we calculated the real part?

Excellent observation. When we multiply the numerator $(x-1 + iy)$ by the conjugate $(2x - i(2y+1))$, we get a complex result. The real part of this product comes from the product of the real parts minus the product of the imaginary parts (because $i^2 = -1$). The terms containing $i$ are the imaginary part, which we ignore because the problem only asks for the real part.

Q: How do I know it's a circle and not a line?

Look at the final equation you derived: $2x^2 + 2y^2 + 2x + 3y + 1 = 0$. A general equation of the form $Ax^2 + Ay^2 + Dx + Ey + F = 0$ represents a circle, provided $A \neq 0$. Since the coefficients of $x^2$ and $y^2$ are equal and non-zero, it is definitely a circle. If the $x^2$ and $y^2$ terms had cancelled out, then it would have been a straight line.

Question

If $Re(\frac{z-1}{2z+i}) = 1$, where $z = x + iy$, then the point $(x, y)$ lies on a

Options:

(A) circle whose centre is at $(-\frac{1}{2}, -\frac{3}{2})$.

(B) straight line whose slope is $\frac{3}{2}$.

(C) circle whose diameter is $\frac{\sqrt{5}}{2}$.

(D) straight line whose slope is $-\frac{2}{3}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the complex plane, a vast, two-dimensional landscape where every point is defined by $z = x + iy$. We are given the condition $Re\left(\frac{z-1}{2z+i}ight) = 1$.

Our first step is to translate the abstract $z$ into the concrete coordinates $(x, y)$. We substitute $z = x + iy$ into the expression:

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

We have a complex fraction, and we must isolate its real part. Since the denominator is complex, we cannot immediately identify the real component.

### The Purification: Rationalizing the Denominator

To see the real part clearly, we must perform a "purification" of the denominator. We multiply both the numerator and the denominator by the complex conjugate of the denominator, which is $2x - i(2y+1)$.

By doing this, we transform the denominator into a purely real value:

$$(2x)^2 + (2y+1)^2 = 4x^2 + 4y^2 + 4y + 1$$

This is the magic of complex conjugates; they turn complex denominators into real ones, effectively clearing the path for us to extract the real part.

### The Extraction: Finding the Real Soul

Now, we focus only on the numerator of our new, rationalized fraction. We need the real part of the product: $((x-1) + iy) \cdot (2x - i(2y+1))$.

When we expand this, we only care about the terms that do not contain $i$. The real part is:

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

The problem states that the real part of the entire fraction is $1$. Therefore, we set our real numerator divided by the real denominator equal to $1$:

$$\frac{2x^2 - 2x + 2y^2 + y}{4x^2 + 4y^2 + 4y + 1} = 1$$

### The Revelation: The Equation of a Circle

Now, we cross-multiply to eliminate the fraction:

$$2x^2 - 2x + 2y^2 + y = 4x^2 + 4y^2 + 4y + 1$$

Moving all terms to one side, we obtain:

$$2x^2 + 2y^2 + 2x + 3y + 1 = 0$$

Dividing by $2$, we arrive at the standard form:

$$x^2 + y^2 + x + \frac{3}{2}y + \frac{1}{2} = 0$$

This is the unmistakable signature of a circle. By comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify the center at $(-\frac{1}{2}, -\frac{3}{4})$.

We calculate the radius $r$ using the formula $r = \sqrt{g^2 + f^2 - c}$:

$$r = \sqrt{\left(\frac{1}{2}ight)^2 + \left(\frac{3}{4}ight)^2 - \frac{1}{2}} = \sqrt{\frac{1}{4} + \frac{9}{16} - \frac{8}{16}} = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}$$

The problem asks for the diameter of this circle. The diameter is $2r$:

$$D = 2 	imes \frac{\sqrt{5}}{4} = \mathbf{\frac{\sqrt{5}}{2}}$$

We have arrived at our destination: a circle with a **diameter of $\frac{\sqrt{5}}{2}$**.

Animated Solution for Mathematics - Complex Numbers: If $R e (\frac{z - 1}{2 z + i}) = 1$ , where $z = x + i y$ , then the point $(x, y)$ lies on a

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Visualized Solution

Visualizing the Complex Point $z$

Substituting $z = x + i y$

Rationalizing the Denominator

Simplifying the Denominator

Extracting the Real Part of the Numerator

Setting the Real Part to $1$

Cross-Multiplying to Simplify

Rearranging into Standard Form

Normalizing the Circle Equation

Finding the Center of the Circle

Calculating the Radius

Final Conclusion: Diameter

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Purification

The Extraction

The Revelation

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