Algebra Of Complex Numbers

We are given that for a complex number . We need to find the maximum value of .

Let's analyze the given condition .

Assume where and is the argument of . Thus, . Then:

Simplifying, we have:

Using the properties of modulus (absolute value), we have:

We want this expression maximized, which means to focus on individual maximizers of both square terms. Notice:

The expression maximizes when each part is in its maximum and when angles related to sine and cosine components fit functionality.

We equate terms for maximum attainable:

Solving , use quadratic formulae structure and trial solutions, which yield in tuning absolute max.

The maximal structure recognizes impact from roots, knowing quadratic form.

Therefore, the maximum value of is , which concurs with option given.

To find where , we follow these steps:

1. First, calculate , which is the complex conjugate of . Thus, .
2. Next, we calculate :
   1. Calculate .
      • Using the formula , we find:
      • since .
      • .
   2. Now find .
      • Using the same expansion formula, .
      • .
      • .
   3. Finally, calculate .
      • Use the distribution method: .
      • .
      • .
      • .
3. Now, calculate , using :
   1. .
   2. .
   3. .
      • Expand: .
      • .
4. Finally, evaluate :
   • Calculate .
   • The imaginary parts cancel out: .

The value of is .

The correct answer is option (A) :

The problem involves finding the minimum possible value of a parameter in a quadratic equation whose roots, and , are natural numbers, subject to certain divisibility conditions on . Once these values are found, we need to evaluate a given expression.

Concept Used:

The solution relies on Vieta's formulas for the roots of a quadratic equation and basic number theory principles (divisibility rules).

For a quadratic equation of the form with roots and , Vieta's formulas state:

• Sum of roots:
• Product of roots:

We will use these relations to connect the roots to the parameter and then apply the given constraints to find the minimum possible value for .

Step-by-Step Solution:

Step 1: Apply Vieta's formulas to the given equation.

The equation is . The roots are and , which are natural numbers ( ).

From Vieta's formulas, we have:

Step 2: Find the minimum possible value for satisfying the given conditions.

The problem states that assumes the minimum possible value under the conditions that and . This means must not be divisible by 2 (i.e., it must be odd) and must not be divisible by 3.

Since and , we can express as a function of one root, say :

To find the minimum value of , we should test values of starting from .

• If , then , and . Here, , so this value is not allowed.
• If , then , and . Here, , so this value is not allowed.
• If , then , and . Here, , so this value is not allowed.
• If , then , and . This is divisible by 2. Not allowed.
• If , then , and .
  • Check divisibility by 2: 325 is odd, so . This condition is satisfied.
  • Check divisibility by 3: The sum of the digits is , which is not divisible by 3. So, . This condition is also satisfied.

Since we are checking values of in increasing order, the first value of that satisfies all conditions will be the minimum possible value. Therefore, the minimum value of is 325, with corresponding roots and .

Step 3: Evaluate the given expression.

The expression to evaluate is:

We have , , and . Let's calculate each part of the expression:

• Numerator Part 1: .
• Numerator Part 2: .
• Denominator: .

Final Computation & Result:

Now, substitute these computed values back into the expression:

The value of the expression is 60.

We are given two equations involving the complex number :

1. The first equation is .
2. The second equation is .

We need to find the number of common roots of these two equations.

Step 1: Analyze the polynomial degree

The first equation is of degree 1985, while the second equation is of degree 3. A polynomial of degree can have at most distinct roots. Therefore, the second equation can have at most 3 roots.

Step 2: Factorization attempt for the second polynomial

Let's factorize the second equation . By using the Rational Root Theorem or trial and error, we can check for possible roots:

Suppose is a root, then substituting into the equation,

Thus, is indeed a root. So, we can factor the polynomial:

Step 3: Solve the quadratic equation

Now solve the quadratic factor . The roots are given by:

Therefore, the roots are:

Step 4: Check common roots

The roots of the original equation are .

Now, for these roots to also satisfy , substitute each root one-by-one into the equation and solve.

• For : Substitute into the first equation,
  
• For and : Since these are complex roots, their magnitudes raise the power and do not satisfy the equation either by trials or roots of unity properties.

Thus, none of the roots of satisfy the equation .

Conclusion

The number of common roots is therefore 0.