Complex Numbers

We are given , and we need to find the imaginary part of .

To simplify the expression for , we multiply both the numerator and the denominator by the conjugate of the denominator, :

First, simplify the denominator:

Now, simplify the numerator:

Thus, we have:

The imaginary part of is .

The answer is .

We are given , and we need to find .

First, find the modulus of . The modulus of a complex number is given by:

For , we have and .

Thus, the modulus of is:

Now, use the property of moduli: for any integer .

Therefore, .

We already know that , s\)

The answer is 49.

We are given that is a root of the quadratic equation:

, where is a real number. Let be the other root, and we need to find .

From Vieta's formulas, for the equation , we know the following:

• (sum of the roots)
• (product of the roots)

Since , we can calculate :

Next, using , we get:

Now, calculate :

Finally, calculate :

The answer is -10.

We are given , and we need to find .

First, multiply both sides of the equation by to find :

Now, distribute :

Since , we have:

Now, we find , the conjugate of :

Now, calculate :

The answer is 26.

We are given three distinct points in the complex plane such that:

• The segment joining and is perpendicular to the segment joining and .
• and .

We need to find .

Since the segment joining and is perpendicular to the segment joining and , we can apply the Pythagorean theorem in the right triangle formed by the points , , .

According to the Pythagorean theorem:

Substituting the given values:

The answer is 13.

We are asked to find the value of the expression:

We can simplify each square root term by recognizing that the square root of a negative number involves the imaginary unit , where

1.

2.

3.

Now, combine all the terms:

The answer is .

We are given the complex number , where and , and we are told that the real part of is . We need to find the value of .

To simplify the expression for , we multiply both the numerator and denominator by the conjugate of the denominator, :

First, simplify the denominator:

Now, simplify the numerator:

Thus, the expression for becomes:

The real part of is given by , and we are told that the real part is :

Now, solve for :

Multiply both sides by :

The answer is .

We are given that the area of the triangle formed by the points on the complex plane is 128, and we need to find .

The vertices of the triangle are represented by the points: , , and .

To find the area of the triangle, we use the formula for the area of a triangle in the complex plane given by the points , , :

Here, the points are , , and , so we need to calculate the complex number\)

The area can also be calculated using the formula for the area of a triangle with vertices at complex numbers , , :

By solving, we find that the area of the triangle is related to the value of , and it is given that the area is 128:

Solving for , we find that .

The answer is 16.

We are given the three vertices of the triangle as the complex numbers , , and . The right angle is at , and the triangle is isosceles, meaning the two other sides are equal in length. To find , we use the distance formula for complex numbers.
1. Calculate the distance between and : The distance between two complex numbers and is given by: Applying this to and , the distance is: 2. Calculate the distance between and : The distance is: The magnitude is: 3. Since the triangle is isosceles, the two distances must be equal: Set the two distances equal to each other: 4. Square both sides to eliminate the square root: 5. Expand both sides: Simplifying: 6. Cancel from both sides and simplify:

The correct option is (B) :

We are given that . First, recall that the conjugate of , denoted , satisfies: Taking the conjugate of both sides of , we get: Thus, we have: Now, the imaginary part of is the coefficient of , which is . Therefore:

The correct option is (A) :

We are given the expression:

Step 1: Simplify the denominator

We start by simplifying the denominator .

- First, multiply : - Now, multiply this result by : Thus, the denominator simplifies to .

Step 2: Simplify the whole expression

Now, simplify the whole expression: Multiply both the numerator and denominator by the conjugate of the denominator to simplify: - The denominator becomes: - The numerator becomes: Thus, the simplified expression is:

Step 3: Find the magnitude

To find the magnitude of this complex number, we calculate: We find that the magnitude is approximately , so the expression simplifies to:

The correct option is (B) :

To minimize , we need to find the minimum sum of the distances from to the points and on the complex plane.

This is a classical problem known as the triangle inequality. The minimum value occurs when lies on the line segment joining and . The distance between and is .
Thus, the minimum value of is , which occurs when lies on the segment between and .

The correct option is (C) : 3

We are given the complex number . To find the principal argument, we first express in polar form. We multiply the numerator and denominator by the conjugate of the denominator to simplify: Now, expand the numerator: Thus, Now, express the complex number in polar form , where Since

The correct option is (B) :

Given that , we know that lies on the unit circle in the complex plane. Also, and are different complex numbers. We need to evaluate the expression .
Let’s consider the geometric interpretation of this expression. The numerator involves , which is a product of the conjugate of and .
The denominator represents the difference between and .
Using properties of complex numbers and their magnitudes, this expression simplifies to 1.
The intuition behind this is that when , the distance between the points corresponding to and on the complex plane is normalized such that the magnitude of the given expression equals 1.

The correct option is (E) :

We are given the complex number . To find the magnitude , we follow these steps: 1. Simplify the expression for : 2. Multiply the numerator : 3. Divide by the denominator . Multiply both the numerator and denominator by the complex conjugate of the denominator, : 4. Simplify the denominator: 5. Simplify the numerator: 6. Therefore, we have: 7. Find the magnitude of , which is :

The correct option is (C) :

Let be a cube root of unity.

Then the cube roots of -2 are , , and .

The expression is

Let . Then the expression becomes .

Let . Then the expression becomes .

Let . Then the expression becomes .

Let . Then the expression becomes .

Let . Then the expression becomes .

Let . Then the expression becomes .

Let . Then the expression becomes .

The expression does not have a single value.

However, -1 is a possible value. For example, if , then we get -1.

Given: has no real roots.

We recall the following properties of real polynomials:

1. Every real polynomial of odd degree has at least one real root (by the Intermediate Value Theorem)

2. Real polynomials of even degree may have no real roots (e.g., )

Since has no real roots, it cannot be of odd degree. Therefore, the degree must be even.

Options analysis: (A) Specific degree (9) - incorrect as it's odd (B) Degree ≤4 - false (e.g., has no real roots) (C) Odd degree - contradicts given condition (D) Degree ≥4 - false (e.g., has no real roots) (E) Even degree - correct

The correct answer is (E) An even number.

Given:

First, rationalize the denominator:

Simplify the denominator:

Thus:

Therefore:

Now compute :

The correct answer is (E) .

Given , we know from the properties of complex numbers that . The expression in question can be rewritten using the properties of the modulus of complex numbers: Using the property of modulus, the value of this expression can be further explored by recognizing it as a special case of the formula for the modulus of a Möbius transformation, where is on the unit circle.
This formula holds true under the condition that the denominator does not become zero, which is when .
If is a unit complex number, the expression simplifies due to the rotational symmetry of complex numbers on the unit circle, maintaining the modulus value:
This is because the transformation preserves the distance ratio due to its conformal nature, keeping the modulus invariant when .

Step 1: We are given that two complex numbers are equal. So, equate the real and imaginary parts of both sides of the equation. Step 2: Simplify both sides: Step 3: Group the real and imaginary terms: Step 4: Equate the real and imaginary parts: 1. 2.
Step 5: Solve the system of equations: From equation 1: From equation 2: Step 6: Solve the system of linear equations. Using equation (1): Substitute in equation (2): Substitute into equation (1): Thus, .

We know that for a complex number to lie on the line , the real part and the imaginary part of the complex number must be equal. Let the complex number be: Multiply both the numerator and denominator by the conjugate of the denominator:
Thus, the real part is and the imaginary part is .
For the complex number to lie on the line , the real part and imaginary part must be equal: Cancel the denominator: Solve for : Thus, the value of is .

A rotation by counterclockwise of a complex number can be achieved by multiplying it by , i.e., the new number is .
So, Thus, the correct answer is (C).

To solve , we can convert the complex number into polar form and then use De Moivre's Theorem. First, express in polar form: So, .
Now, apply De Moivre's Theorem to : Simplifying the powers and the angles:
Thus, , and .
Therefore: So the correct answer is (B) .