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JEE Main 2024
Complex Numbers
Complex Numbers
Hard

Question

If the set R={(a,b):a+5b=42,a,bN}R=\{(a, b): a+5 b=42, a, b \in \mathbb{N}\} has mm elements and \sum_\limits{n=1}^m\left(1-i^{n !}\right)=x+i y, where i=1i=\sqrt{-1}, then the value of m+x+ym+x+y is

Options

Solution

Key Concepts and Formulas

  • Diophantine Equations: Solving linear equations of the form ax+by=cax + by = c for integer solutions, often with the constraint that the solutions are natural numbers.
  • Properties of Complex Numbers: Understanding the cyclic nature of powers of ii, where i1=ii^1 = i, i2=1i^2 = -1, i3=ii^3 = -i, and i4=1i^4 = 1. This cycle repeats every four powers.
  • Factorials and Modular Arithmetic: Understanding that for n4n \ge 4, n!n! is divisible by 4, meaning n!0(mod4)n! \equiv 0 \pmod{4}.

Step-by-Step Solution

Step 1: Determine the number of elements (mm) in the set R

The set RR is defined as R={(a,b):a+5b=42,a,bN}R = \{(a, b) : a + 5b = 42, a, b \in \mathbb{N}\}. We need to find the number of pairs (a,b)(a, b) of natural numbers that satisfy the equation a+5b=42a + 5b = 42.

Step 2: Express aa in terms of bb

We rewrite the equation to express aa as a function of bb: a=425ba = 42 - 5b

Step 3: Apply the natural number constraint for aa

Since aNa \in \mathbb{N}, we must have a1a \ge 1. Substituting the expression for aa, we get: 425b142 - 5b \ge 1 415b41 \ge 5b b415b \le \frac{41}{5} b8.2b \le 8.2

Step 4: Apply the natural number constraint for bb

Since bNb \in \mathbb{N}, we must have b1b \ge 1.

Step 5: Find all possible integer values for bb

Combining the constraints b1b \ge 1 and b8.2b \le 8.2, the possible integer values for bb are {1,2,3,4,5,6,7,8}\{1, 2, 3, 4, 5, 6, 7, 8\}.

Step 6: Calculate the corresponding aa values for each bb

We iterate through each valid bb and find the corresponding aa:

  • If b=1b = 1, a=425(1)=37a = 42 - 5(1) = 37. So, (37,1)R(37, 1) \in R.
  • If b=2b = 2, a=425(2)=32a = 42 - 5(2) = 32. So, (32,2)R(32, 2) \in R.
  • If b=3b = 3, a=425(3)=27a = 42 - 5(3) = 27. So, (27,3)R(27, 3) \in R.
  • If b=4b = 4, a=425(4)=22a = 42 - 5(4) = 22. So, (22,4)R(22, 4) \in R.
  • If b=5b = 5, a=425(5)=17a = 42 - 5(5) = 17. So, (17,5)R(17, 5) \in R.
  • If b=6b = 6, a=425(6)=12a = 42 - 5(6) = 12. So, (12,6)R(12, 6) \in R.
  • If b=7b = 7, a=425(7)=7a = 42 - 5(7) = 7. So, (7,7)R(7, 7) \in R.
  • If b=8b = 8, a=425(8)=2a = 42 - 5(8) = 2. So, (2,8)R(2, 8) \in R.

Since all pairs have natural numbers for both aa and bb, the set RR is {(37,1),(32,2),(27,3),(22,4),(17,5),(12,6),(7,7),(2,8)}\{(37, 1), (32, 2), (27, 3), (22, 4), (17, 5), (12, 6), (7, 7), (2, 8)\}.

Step 7: Determine mm

The number of elements in set RR is m=8m = 8.

Step 8: Evaluate the summation n=1m(1in!)\sum_{n=1}^m (1 - i^{n!})

We need to evaluate the sum n=18(1in!)=x+iy\sum_{n=1}^8 (1 - i^{n!}) = x + iy. We can split the sum: n=18(1in!)=n=181n=18in!\sum_{n=1}^8 (1 - i^{n!}) = \sum_{n=1}^8 1 - \sum_{n=1}^8 i^{n!}

Step 9: Calculate n=181\sum_{n=1}^8 1

The first part is n=181=1×8=8\sum_{n=1}^8 1 = 1 \times 8 = 8.

Step 10: Analyze in!i^{n!} for each value of nn from 1 to 8

The value of iki^k depends on k(mod4)k \pmod{4}.

  • For n=1n = 1: n!=1!=1n! = 1! = 1. i1!=i1=ii^{1!} = i^1 = i.
  • For n=2n = 2: n!=2!=2n! = 2! = 2. i2!=i2=1i^{2!} = i^2 = -1.
  • For n=3n = 3: n!=3!=6n! = 3! = 6. 6(mod4)26 \pmod{4} \equiv 2. i3!=i6=i2=1i^{3!} = i^6 = i^2 = -1.
  • For n=4n = 4: n!=4!=24n! = 4! = 24. 24(mod4)024 \pmod{4} \equiv 0. i4!=i24=1i^{4!} = i^{24} = 1.
  • For n4n \ge 4: For any integer n4n \ge 4, the term n!n! will always contain 44 as a factor. This means n!n! is always a multiple of 44. Therefore, for n=4,5,6,7,8n = 4, 5, 6, 7, 8, n!0(mod4)n! \equiv 0 \pmod{4}, which implies in!=1i^{n!} = 1.

Step 11: Calculate the sum n=18in!\sum_{n=1}^8 i^{n!}

n=18in!=i1!+i2!+i3!+i4!+i5!+i6!+i7!+i8!\sum_{n=1}^8 i^{n!} = i^{1!} + i^{2!} + i^{3!} + i^{4!} + i^{5!} + i^{6!} + i^{7!} + i^{8!} =i+(1)+(1)+1+1+1+1+1 = i + (-1) + (-1) + 1 + 1 + 1 + 1 + 1 =i2+5 = i - 2 + 5 =3+i = 3 + i

Step 12: Calculate the total sum n=18(1in!)\sum_{n=1}^8 (1 - i^{n!})

n=18(1in!)=(n=181)(n=18in!)\sum_{n=1}^8 (1 - i^{n!}) = \left( \sum_{n=1}^8 1 \right) - \left( \sum_{n=1}^8 i^{n!} \right) =8(3+i) = 8 - (3 + i) =83i = 8 - 3 - i =5i = 5 - i

Step 13: Identify xx and yy

We have n=1m(1in!)=x+iy\sum_{n=1}^m (1 - i^{n!}) = x + iy. Comparing 5i5 - i with x+iyx + iy, we get x=5x = 5 and y=1y = -1.

Step 14: Calculate m+x+ym + x + y

Substitute the values we found: m=8m = 8, x=5x = 5, and y=1y = -1. m+x+y=8+5+(1)m + x + y = 8 + 5 + (-1) m+x+y=131m + x + y = 13 - 1 m+x+y=12m + x + y = 12

Common Mistakes & Tips

  • Natural Number Definition: Always remember that natural numbers start from 1 (i.e., N={1,2,3,...}\mathbb{N} = \{1, 2, 3, ...\}).
  • Powers of i: Be careful when evaluating powers of ii. Remember the cycle: i1=i,i2=1,i3=i,i4=1i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. Also, remember to use modulo 4.

Summary

By carefully analyzing the Diophantine equation and the properties of complex numbers, we found m=8m=8, x=5x=5, and y=1y=-1. Therefore, m+x+y=12m+x+y = 12.

The final answer is \boxed{12}, which corresponds to option (A).

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