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The equation \(\operatorname{Im}\left(\frac{i z-2}{z-i}\right)+1=0, z \in C, z \neq i\) represents a part of a circle having radius equal to : (a) 2 (b) 1 (c) \(\frac{3}{4}\) (d) \(\frac{1}{2}\)

Short Answer

Expert verified
The radius of the circle is 1.

Step by step solution

01

Understand the Problem

We are given the complex equation \(\operatorname{Im}\left(\frac{i z-2}{z-i}\right)+1=0, z \in \mathbb{C}, z eq i\). This means the imaginary part of the expression is -1. We need to identify which part of the resulting geometric figure represents a circle and find its radius.
02

Simplify Imaginary Part

Let \(w = \frac{i z - 2}{z - i}\). Then the equation becomes \(\operatorname{Im}(w) = -1\). Express \(z\) as \(z = x + yi\). Substitute and separate real and imaginary parts before finding \(\operatorname{Im}(w)\).
03

Substitute and Simplify

Substitute \(z = x + yi\) into the equation, giving \(w = \frac{i(x + yi) - 2}{(x + yi) - i}\). Simplify the expression: \(w = \frac{(-y + x)i - 2}{(x - 1) + yi}\). Multiply numerator and denominator by conjugate \((x - 1) - yi\).
04

Calculate Imaginary Part

After simplification, calculate the imaginary component of \(w\):\[ \operatorname{Im}(w) = \frac{-2y + x^2 - x + y^2 - 1}{(x - 1)^2 + y^2} \]By setting \(\operatorname{Im}(w) = -1\), simplify to find:\[ -2y + x^2 - x + y^2 - 1 = -(x - 1)^2 - y^2 \]
05

Solve the Radius

Set the imaginary part equal to \(-1\):\[ x^2 - x + y^2 - 2y = -1((x-1)^2 + y^2) + 1 \]Simplifying further reveals:\[ x^2 - x + y^2 - 2y = -x^2 + 2x - 1 - y^2 \]Combine terms, simplify and arrange into circle equation form:\[ (x - \frac{3}{2})^2 + (y - 1)^2 = 1 \]
06

Identify the Radius

The equation \((x-\frac{3}{2})^2 + (y-1)^2 = 1\) is a circle centered at \((\frac{3}{2}, 1)\) with radius \(1\). Hence, using geometry of the equation, the radius is calculated from the simplified form indicating the equation represents a circle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Imaginary Part
Understanding complex numbers involves dissecting both their real and imaginary parts. The imaginary part is crucial because it helps in visualizing complex numbers in the complex plane. A complex number is represented as \( z = x + yi \), where \( x \) and \( y \) are real numbers and \( i \) is the imaginary unit \( i = \sqrt{-1} \). Here, \( y \) is the imaginary part, which cannot be directly observed but helps in complex calculations and visual representation.

In the context of the given problem, which involves the complex equation \( \operatorname{Im}\left(\frac{i z-2}{z-i}\right)+1=0 \), isolating the imaginary part of the expression \( \frac{i z-2}{z-i} \) is vital. It essentially involves mathematical manipulation by separating real and imaginary components after substituting \( z = x + yi \). Thus, the challenge lies in simplifying the fraction and ensuring the imaginary component matches the required value, making it robust for identifying geometric shapes.
Circle Equations
Equations of a circle are a profound part of geometry and an essential concept when working with complex numbers on the Cartesian plane. Typically, a circle equation in the plane is given by \((x - h)^2 + (y - k)^2 = r^2\), where \( (h, k) \) denotes the circle's center, and \( r \) is the radius.

When dealing with complex numbers, these equations often arise when extracting the real and imaginary components of a number or expression, as seen in our exercise. Here, substituting the complex variable and simplifying leads us to an equation: \( (x - \frac{3}{2})^2 + (y - 1)^2 = 1 \). This is standard form and represents a circle with a center at \( (\frac{3}{2}, 1) \) and a radius of 1, making it easier to visualize in both algebraic and geometric contexts.

Having this understanding allows learners to relate complex number properties with geometric shapes, offering a dual perspective from algebra and geometry.
Geometric Representation
Geometric representation of complex numbers transforms algebraic expressions into visual aids, providing intuitive and easy-to-understand insights into complex numbers. In the case of \( z = x + yi \), the geometric representation would be a point in the complex plane with coordinates \( (x, y) \). This view simplifies many operations, as it allows for straightforward assessment of distances and relationships between numbers.

For equations, such as the one in the exercise, manipulating and simplifying the imaginary part can lead to a geometric shape, like a circle. The equation \( (x - \frac{3}{2})^2 + (y - 1)^2 = 1 \), for example, defines a specific position and size of a circle within the plane.

Understanding this representation helps in visualizing problems, seeing beyond numerical values to the shapes and patterns they form. This skill is particularly useful in physics, engineering, and computer science, where complex numbers frequently model real-world phenomena.

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Most popular questions from this chapter

Let \(\mathrm{A}(4,-4)\) and \(\mathrm{B}(9,6)\) be points on the parabola, \(y^{2}=4 x\). Let \(\mathrm{C}\) be chosen on the arc \(\mathrm{AOB}\) of the parabola, where \(\mathrm{O}\) is the origin, such that the area of \(\triangle \mathrm{ACB}\) is maximum. Then, the area (in sq. units) of \(\triangle \mathrm{ACB}\), is: (a) \(31 \frac{1}{4}\) (b) \(30 \frac{1}{2}\) (c) 32 (d) \(31 \frac{3}{4}\)

The tangent to the circle \(C_{1}: x^{2}+y^{2}-2 x-1=0\) at the point (2, 1) cuts off a chord of length 4 from a circle \(C_{2}\) whose centre is \((3,-2)\). The radius of \(C_{2}\) is (a) \(\sqrt{6}\) (b) 2 (c) \(\sqrt{2}\) (d) 3

The locus of the centres of the circles, which touch the circle, \(x^{2}+y^{2}=1\) externally, also touch the \(y\)-axis and lie in the first quadrant, is: (a) \(x=\sqrt{1+4 y}, y \geq 0\) (b) \(y=\sqrt{1+2 x}, x \geq 0\) (c) \(y=\sqrt{1+4 x}, x \geq 0\) (d) \(x=\sqrt{1+2 y}, y \geq 0\)

If \(y+3 x=0\) is the equation of a chord of the circle, \(x^{2}+y^{2}-30 x=0\), then the equation of the circle with this chord as diameter is: (a) \(x^{2}+y^{2}+3 x+9 y=0\) (b) \(x^{2}+y^{2}+3 x-9 y=0\) (c) \(x^{2}+y^{2}-3 x-9 y=0\) (d) \(x^{2}+y^{2}-3 x+9 y=0\)

Consider a family of circles which are passing through the point \((-1,1)\) and are tangent to \(x\)-axis. If \((h, k)\) are the coordinate of the centre of the circles, then the set of values of \(k\) is given by the interval (a) \(-\frac{1}{2} \leq k \leq \frac{1}{2}\) (b) \(k \leq \frac{1}{2}\) (c) \(0 \leq k \leq \frac{1}{3}\) (d) \(k \geq \frac{1}{2}\)

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