Sigma Percentile
JEE Main 2020 (9 January Shift 1)
LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: Let be a complex number such that and . Then the value of is:

Select Answer:

Visualized Solution

Initial Setup in Argand Plane

  • Given: and
  • Objective: Find the value of
  • Let be the complex number.

Locus of

  • Rewrite as:
  • Geometrically, is equidistant from points and .
  • This represents the perpendicular bisector of the segment joining these points.

Solving for the Imaginary Part

  • Substitute :
  • Expand:
  • Simplify:

Modulus Condition

  • Given:
  • Squaring both sides:
  • In Cartesian form:
  • This represents a circle centered at the origin with radius .

Calculating the Real Part

  • Substitute into the circle equation:

Defining the Target

  • We need to find .
  • Geometrically, this is the distance between and the point .
  • Using distance formula:

Substituting and

  • Substitute and :
  • Simplify inside the bracket:

Final Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

Imagine you are standing on the Argand plane, a vast, two-dimensional landscape where every point is a complex number . We are embarking on a journey to find the hidden location of a point that obeys two strict, beautiful laws.
Our first law is given by the equation:
By cross-multiplying, we transform this into . This is a profound geometric statement; it tells us that the distance from to the point (which is ) is exactly equal to the distance from to the point (which is ).

The Line of Symmetry

The locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them. The midpoint between and is , and since the segment is vertical, the perpendicular bisector must be a horizontal line.
Algebraically, if we set , the equation becomes:
Watch as the terms vanish, leaving us with . Simplifying this, we find , or:
This horizontal line is the home of our complex number .

The Circle of Constraints

Now, we introduce our second law: . This is the equation of a circle centered at the origin with a radius of .
In Cartesian terms, this is:
We now have two constraints: our point must lie on the line and on the circle . By substituting into the circle equation, we get:
This simplifies to , which leads us to .

The Final Destination

The question asks for the value of . Geometrically, this is the distance between our point and the point .
Using the distance formula, we have:
We already know and . Substituting these values, we get:
Simplifying the term inside the parenthesis, we have . Squaring this gives .
Finally, we calculate:
The final result is .

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