Complex Numbers

The correct answer is 2
Let z = x + iy
So 2x = (1 + i)(x² – y² + 2xyi)
⇒ 2x = x² – y² – 2xy …(i) and x² – y² + 2xy = 0 …(ii)
From (i) and (ii) we get
x = 0 or y
When x = 0 we get y = 0
When y
we get

So there will be total 3 possible values of z, which are
and
Sum of squares of modulus

= 2

The correct answer is (C) : 2
C₁ : |z – (4 + 3i)| = 2 and C₂ : |z| + |z – 4| = 6, z ∈ C
C₁: represents a circle with centre (4, 3) and radius 2 and C₂ represents a ellipse with focii at (0, 0) and (4, 0) and length of major axis = 6, and length of semi-major axis 2√5 and (4, 2) lies inside the both C₁ and C₂ and (4, 3) lies outside the C₂

Therefore,
no. of intersection points = 2

The correct answer is (C) : 1
is purely imaginary



and

is purely real.




= 1-i and = -1-i



= 1

The correct answer is (C) : nowhere

|z| = 3
arg(z-1) - arg(z+1) = π/4

⇒ LK = AL = α = 1
K(0,1)
Radius =
PL = PK + KL =
P(0,1 + )
Therefore , Number of intersection = 0

The correct answer is (C) : x + 2y + 4 = 0
|z-i| = |z+5i|
Thereafter, z lies on ⊥r bisector of (0, 1) and (0, –5)
i.e., line y = –2
as |z| = 2
⇒ z = –2i
x = 0 and y = –2Hence, x + 2y + 4 = 0

The correct answer is (A) : 1
1 + x² + x⁴ = 0
ω¹⁰¹¹ + ω²⁰²² - ω³⁰³³
= (ω³)³³⁷ + (ω³)⁶⁷⁴ - (ω³)¹⁰¹¹
= 1+1-1
= 1

The correct answer is 80
Here |z – 3| < 1

And

∴ Coordinate of P = (3 – cosθ, sinθ)

The correct answer is 2
∵ z² + z + 1 = 0
⇒ ω or ω²

=
= |0 + 0 – 2|
= 2

To solve the problem, we need to find all complex numbers such that .
Let , where and is the imaginary unit. The conjugate .

Substitute into the equation:

Expanding :
.
Therefore, the equation becomes:
.
Separate the real and imaginary parts:
.
Since the equation must hold for both real and imaginary parts:

1. (Equation 1)
2. (Equation 2)

From Equation 2, factor out :
.
So, or .

If , from Equation 1:

, leading to or .
Thus, possible are and (as these have ).

If , then . Substituting in Equation 1:



. Thus, additional solutions are and .

So . Calculate :

- For , .
- For , .
- For , .
- For , .

Summing these:
.

The computed sum is , which fits within the range [0, 0].

The correct answer is (B) : A portion of a circle centred at that lies in the second only

and
is a part of circle as shown

The correct answer is (C) :
Let P = (2λ+1,λ+3,2λ+2)
Let Q =


and
PQ =

The correct answer is (C) : 496




at x=−1

Given the condition:

Let , where and are real numbers. Substituting into the equation:

Squaring both sides to eliminate the square roots:

Cross-multiplying:

Expanding both sides:

Bringing all terms to one side:

To express this in standard circle form, complete the square for the -terms:

This represents a circle with center at and radius 2.

Thus, the correct answer is option (4)

We are tasked with analyzing the given inequalities for the statements and involving complex numbers and .

Step 1: Analyze the Inequality

The inequality is given as:

Expanding both sides:

The inequality becomes:

Step 2: Check Statement

For , we are given:

Substitute these values into the inequality:

This inequality is not valid. Hence, is false.

Step 3: Check Statement

For , the inequality is:

Expanding:

For , we are given:

Substitute these values:

This inequality is not valid. Hence, is false.

Final Answer:

Both and are false.

We are tasked with solving the given expression:

Step 1: Simplify the condition

Rewrite the given condition:

Let:

Step 2: Imaginary part condition

The imaginary part of the denominator must satisfy:

is imaginary. Substitute

so:

For the expression to be purely imaginary:

Step 3: Solve the real part equation

Rewriting:

Factorize:

Step 4: Solve for

Let:

where:

Substitute and into

Simplify:

Step 5: Calculate

Substitute

Simplify:

Final Answer:

Let , where . Then, . Given the equation: Substitute and : Expand : Substitute back: Simplify: Equate real and imaginary parts: Solve :
Case 1: Substitute into the imaginary part: So, or . Thus, or .
Case 2: Substitute into the imaginary part: Solve the quadratic equation: So, or . Thus, or . The set is: Calculate for each : Sum of : Therefore, the sum is equal to .

We are given the complex number and the equation:

where denotes the complex conjugate of . We are tasked to find .

Step 1: Compute the complex conjugate of

The complex conjugate of is:

Step 2: Simplify the numerator

The numerator is . Substitute :

Expand the product:

Thus, the numerator becomes:

Step 3: Simplify the denominator

The denominator is . Substitute and :

Expand the product:

Step 4: Simplify

Substitute the simplified numerator and denominator into the expression for :

To simplify further, multiply the numerator and denominator by the complex conjugate of the denominator, :

Expand the denominator:

Expand the numerator:

Thus:

Step 5: Find the argument of

The complex number has real part and imaginary part . The argument is given by:

Since , we have:

Step 6: Compute

Finally, multiply the argument by :

Conclusion

The value of is:

To determine the minimum value of , given that the complex number satisfies , we will proceed as follows:

The expression can be rewritten using a substitution. Let the complex number be represented as , where and are real numbers.

The given condition translates to the inequality:

Now, rewrite the target expression:

This simplifies to:

According to the properties of complex numbers, the modulus is given by:

We seek to minimize this expression under the constraint .

Geometrical Interpretation:

The expression represents the distance from the point to any point on or outside the circle centered at the origin with radius 1, represented by .

The shortest distance from the point to the circle centered at the origin occurs along the line passing through the origin and . We find this distance by first calculating the distance from the origin to the point :

The closest point on the circle to would achieve this minimum distance reduced by the circle's radius, which is 1:

Thus, the minimum value of under the condition is .

Therefore, the correct answer is: .

Step 1: Find the roots of the quadratic equation.

Hence, Since , we take:

Step 2: Use relations between roots.

From the equation :

Step 3: Express higher powers of and .

Since is a root: We will use this recurrence relation:

Step 4: Compute successive powers.

Substitute :

Substitute :

Substitute :

Substitute :

Step 5: Compute similarly.

Since satisfies the same relation :

Step 6: Substitute in the given expression.

We need: Substitute the derived values:

Now: Simplify: \]

Step 7: Use

Final Answer:

Express as a Complex Number:
Let where and .

Given:
Here, , so the equation becomes:

Separating Real and Imaginary Parts:
The real part:

The imaginary part:
From the imaginary part, either or .

Case 1:
Substituting into the real part:
This gives or . Since we are looking for non-zero solutions, we have:

Case 2:
Substituting into the real part: which gives no non-zero solutions for .

Solutions:
Thus, the non-zero solutions are and .

Calculating and :

Computing :

Given

Let satisfy . Then the roots are cube roots of unity (other than 1): or with .

Goal

Express in the form and compute

Solution

Choose . Using : Since , we have . Hence Therefore, when written as with , So

Given the equation of the circle and the region in the complex plane:

and the inequality representing the region:

Now, let the points in the shaded region be represented by coordinates .

Possible points are:

Compute the modulus squared for each point:

Hence,

Therefore, the sum is:

To solve the given problem, we start by examining the equation:

This equation essentially represents a condition involving the distances between two points on the complex plane.

First, let's simplify the equation:

Consider as the distance from to :

- If we assume that lies on a circle of radius 1, then we have:

Similarly, consider the absolute value:

represents the modulus of a complex number.

- If we assume lies on a circle of radius , this satisfies the equation provided since both sides could balance out to remain equal.

Therefore, the correct interpretation is that either:

• lies on a circle of radius 1.
• or lies on a circle of radius .

Conclusion: The correct answer is that either lies on a circle of radius 1 or lies on a circle of radius .

Given:
The following series of calculations are provided. Let's go step by step:

Step 1:

Step 2:
We have , which simplifies to .

Step 3:
Next, we calculate .

Step 4:
For the distance between point (1, 4) and the line , we use the distance formula:

Substituting and point (x₁, y₁) = (1, 4), we calculate:

Final Answer: The distance from the point (1, 4) to the line is .

To determine the modulus and amplitude of the complex number , we proceed as follows:

The given complex number is in the form , where and .

To find the modulus of , use the formula:

Substitute and :

Using the identity :

Next, calculate the amplitude using the formula :

because the complex number is in the second quadrant.

Thus, the pair is .

The correct answer is therefore .

We start with the given condition:

Let . Then,

Rationalizing the denominator:

Now, the real part of the expression is:

This implies:

Hence, the equation represents a circle with center at the origin and radius 2.

To find the maximum value of :

This represents the maximum distance of the point from the circle .

Let the center of the circle be and radius .

The distance from to is:

Therefore, the maximum distance is:

Final Answer:

Step 1: Find the roots of the quadratic equation.

The given quadratic equation is: The roots are given by the quadratic formula: Thus, the roots are: So, we have: since the imaginary part of is positive.

Step 2: Express the powers of and .

We are given the equation: We first recognize that both and are complex conjugates, and we use their polar form. Let , and we express it in polar form: The argument of is: Thus, we write: Similarly, for , we have: We now find and : Therefore: This expression matches the given equation: From this, we can equate the real and imaginary parts to find .

Step 3: Find .

After solving the system of equations using the above approach, we get: Thus:

Final Answer: