Sigma Percentile
JEE Main 2019 (9 January)
LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: Let and be two roots of the equation , then is equal to :

Select Answer:

Visualized Solution

Analyze the Given Equation

  • Given equation:
  • Roots are and .
  • Target: Find the value of .

Completing the Square

  • Rewrite the equation:
  • This simplifies to:
  • Rearranging:

Finding the Roots and

  • Taking square root:
  • Roots:
  • Let and .

Calculating and

Raising to the Power of

  • Since
  • Similarly,

Simplifying and

Final Summation

  • Sum
  • Sum
  • Sum
  • Final Answer: -256

The Sigma Insight: Algebraic Operations on Complex Numbers

Solution Diagram

Analyzing the Setup

Imagine you are standing before the quadratic equation . At first glance, it looks like a standard, unassuming problem.
However, as an aspirant, you know that appearances can be deceiving. If you dive straight into the quadratic formula, you will find yourself wading through a sea of calculations. There is a more elegant path, a shortcut that reveals the hidden symmetry of the complex plane.

The Art of Completing the Square

Instead of brute force, let us look at the structure. We can rewrite the equation as .
Because is a perfect square, this transforms our equation into:
This simplifies to . Suddenly, the path clears as we step into the realm of complex numbers. Taking the square root of both sides, we find , which gives us our roots:

The Power of Squaring

Now, the challenge is to find . If you try to expand using the binomial theorem, you will be trapped in a labyrinth of terms.
Let us be smarter and square these roots first. For , we have:
Since , the real parts cancel out, leaving us with . Similarly, for :
This is the turning point. Squaring has simplified our complex numbers into pure imaginary numbers. Geometrically, you have just doubled the angle of these roots on the Argand plane and squared their magnitudes.

The Leap to the Fifteenth Power

We need the fifteenth power, but we have the squares. Let us aim for the fourteenth power first:
This expands to . We know that , and since , we have .
Thus, . By the same logic:

The Final Convergence

We are almost at the finish line. To find , we simply multiply by :
Similarly, for :
Now, for the grand finale, we add them together:
The imaginary terms and vanish into thin air. This leaves us with the final result:
This is the beauty of mathematics. By choosing the right perspective—by seeing the structure rather than just the numbers—we turned a daunting calculation into a graceful dance of cancellation.

Similar Questions

JEE Main 2022 (29 June Shift 1)
LEVELJEE Main

Let and be the roots of the equation . Then, the value of is equal to :

(A)
50
(B)
250
(C)
1250
(D)
1500
JEE Main 2025 (January)
LEVELJEE Main

If and are the roots of then is equal to :

(A)
-2
(B)
6
(C)
-6
(D)
2
JEE Main 2024 (04 Apr Shift 1)
LEVELJEE Main

Let and be the sum and the product of all the non-zero solutions of the equation . Then is equal to :

(A)
6
(B)
8
(C)
2
(D)
4
JEE Main 2019 (8 April Shift 1)
LEVELJEE Main

If and be the roots of the equation , then the least value of for which is :

(A)
(B)
(C)
(D)
JEE Main 2021 (25 February Shift 2)
LEVELJEE Main

If are such that (here ) is a root of , then is equal to:

(A)
7
(B)
-3
(C)
3
(D)
-7
JEE Main 2025 (January)
LEVELJEE Main

If and are the roots of the equation where , then is equal to

(A)
441
(B)
398
(C)
312
(D)
409
JEE Main 2023 (08 Apr Shift 1)
LEVELJEE Main

If for , , then and are the roots of the equation

(A)
(B)
(C)
(D)
JEE Main 2024 (29 Jan Shift 1)
LEVELJEE Main

If , is such that and , then is equal to

(A)
-4
(B)
3
(C)
2
(D)
-1
JEE Advanced 2024
LEVELJEE Main

Let be a polynomial with real coefficients such that . Suppose that is a root of the equation , where . If , and are all the roots of the equation , then is equal to ________.

JEE Main 2024 (31 Jan Shift 2)
LEVELJEE Main

Let and be two complex number such that and . Then equals

(A)
(B)
75
(C)
(D)