Problem 116
Question
If the complex numbers \(z_{1}, z_{2}, z_{3}\) are the vertices \(A\), \(B, C\) respectively of an isosceles right angled triangle with right angle at \(C\), then \(\left(z_{1}-z_{2}\right)^{2}=k\left(z_{1}-z_{3}\right)\left(z_{3}-z_{2}\right)\), where \(k=\) (A) 1 (B) 2 (C) 4 (D) None of these
Step-by-Step Solution
Verified Answer
The value of \(k\) is 2.
1Step 1: Understanding the Isosceles Right Triangle
In an isosceles right triangle, the two legs are equal, and the hypotenuse is equal to \( \sqrt{2} \) times the length of each leg. Let the complex numbers \( z_1 = x_1 + iy_1 \), \( z_2 = x_2 + iy_2 \), and \( z_3 = x_3 + iy_3 \) denote points in the complex plane constituting vertices of this triangle. We have to determine the relation using these vertices.
2Step 2: Determine Relation from Right Triangular Property
In such a triangle, the square of the length of the hypotenuse (diagonal) is equal to the sum of the squares of the lengths of the other two sides (legs). So, we have the condition: \[ |z_{1} - z_{2}|^2 = |z_{1} - z_{3}|^2 + |z_{3} - z_{2}|^2 \].And since it's an isosceles triangle, \(|z_1 - z_3| = |z_3 - z_2| \).
3Step 3: Apply the Pythagorean Theorem
We recognize that the triangle can be viewed as a set of complex numbers where:\[ |z_{1} - z_{2}|^2 = 2|z_{1} - z_{3}|^2 \].This arises because both legs being equal implies each squared, added and multiplied by 2, equals the square of the hypotenuse.
4Step 4: Connect to Given Equation with Complex Numbers
From the geometry, using equality of legs, the algebra becomes:\[ (z_1 - z_2)^2 = 2 (z_1 - z_3)(z_3 - z_2) \].This holds as the expression derived is consistent with the problem's given format, demonstrating the use of an angle \(90^\circ\) at \(C\) creating two equal legs and their squared forms in terms of \( (z_1-z_3) \) and \( (z_3 - z_2) \).
5Step 5: Identify the Correct 'k' Value
Comparing directly with the original assertion given in the problem, the constant \(k\), representing the squared scaling factor due to the right-angle vertex manipulation within the complex form is 2 following from the previous step. Hence, \(k = 2\).
Key Concepts
Isosceles Right TriangleGeometry in Complex PlanePythagorean Theorem
Isosceles Right Triangle
An isosceles right triangle is a type of right triangle that has two sides of equal length, called legs, and one side that is longer, called the hypotenuse. The properties of this triangle make it unique and easy to recognize:
- The two legs are equal in length.
- The hypotenuse is precisely \( \sqrt{2} \) times the length of each leg. This relationship can be derived using the Pythagorean theorem.
- The angles adjacent to the hypotenuse are each \( 45^\circ \), meaning they sum up to \( 90^\circ \) at the base.
Geometry in Complex Plane
In mathematics, complex numbers provide a powerful way of analyzing geometric problems by combining the axes of a traditional coordinate plane into a single complex plane. Each point is represented by a complex number combining real and imaginary parts, \( z = x + iy \), where \( x \) is the real part and \( y \) is the imaginary part.
When dealing with complex numbers in geometric problems, such as triangles, the following points are crucial:
When dealing with complex numbers in geometric problems, such as triangles, the following points are crucial:
- Distance between two complex numbers \( z_1 \) and \( z_2 \) is given by \( |z_1 - z_2| \), similar to finding the length of a segment between two points in the Cartesian plane.
- The concept of rotation and reflection can be easily handled using complex multiplication, which accounts for angles between lines in geometric figures.
- Angular relationships and symmetries are captured through the properties of complex conjugates and magnitudes.
Pythagorean Theorem
The Pythagorean theorem is a timeless principle in geometry, particularly useful in right triangles. It states that in a right triangle, the square of the hypotenuse's length (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In equation form, this is written as:
\[ c^2 = a^2 + b^2 \] Where \( c \) represents the hypotenuse and \( a \) and \( b \) are the triangle's legs.
In the context of an isosceles right triangle:
\[ c^2 = a^2 + b^2 \] Where \( c \) represents the hypotenuse and \( a \) and \( b \) are the triangle's legs.
In the context of an isosceles right triangle:
- Both legs \( a \) and \( b \) are equal, so the formula becomes \( c^2 = 2a^2 \).
- This simplifies to \( c = a\sqrt{2} \), illustrating the fixed relationship of side lengths in such a triangle.
- This theorem finds extensive application in the complex plane, as demonstrated in the original exercise, allowing one to express distances and verify equality relations among sides.
Other exercises in this chapter
Problem 111
If \(|z| \leq 1,|\omega| \leq 1\), then \(|\mathrm{z}-\omega|^{2}\) \((\mathrm{A}) \leq(|z|-|\omega|)^{2}-(\operatorname{Arg} z-\operatorname{Arg} \omega)^{2}\)
View solution Problem 113
If \(k=\frac{3 n}{2}\), where \(n\) is an even positive integer, then \(\sum_{r=1}^{k}(-3)^{r-1} \cdot{ }^{3 n} C_{2 r-1}=\) (A) 0 (B) 1 (C) \(-1\) (D) None of
View solution Problem 118
If \(2 \sqrt{2 x^{4}}=(\sqrt{3}-1)+i(\sqrt{3}+1)\), then \(x=\cos \frac{1}{4}(2 n \pi+k)+i \sin \frac{1}{4}(2 n \pi+k) ;\) \(n=0,1,2,3\), where \(k=\) (A) \(\fr
View solution Problem 119
\(\sum_{p=1}^{32}(3 p+2)\left[\sum_{q=1}^{10}\left(\sin \frac{2 q \pi}{11}-i \cos \frac{2 q \pi}{11}\right)\right]^{p}=\) (A) \(8(1-i)\) (B) \(16(1-i)\) (C) \(4
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