Problem 187
Question
Assertion: If the area of the triangle on the arganc plane formed by the complex numbers \(-z, i z, z-i z\) is 600 square units, then \(|z|=20\) Reason: Area of the triangle on the argand plane formed by the complex numbers \(-z, i z, z-i z\) is \(\frac{3}{2}\left|z^{2}\right|\)
Step-by-Step Solution
Verified Answer
The modulus \\( |z| = 20 \\).
1Step 1: Understand the Triangle Formulation
We are given three vertices of a triangle in the Argand plane: - Vertex 1 at \(-z\).- Vertex 2 at \(iz\).- Vertex 3 at \(z - iz\).
2Step 2: Complex Number to Coordinates Conversion
Convert each complex number to corresponding coordinates:- \(-z\) is \((-x, -y)\).- \(iz\) is \((-y, x)\).- \(z - iz\) is \(x + y, y - x\).
3Step 3: Use the Area of Triangle Formula
The formula for the area of a triangle formed by three points \( (x_1,y_1), (x_2,y_2), (x_3,y_3) \) is given by:\[\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\]
4Step 4: Substitute Coordinates into Area Formula
Substitute the coordinates of the points into the area formula:\[\text{Area} = \frac{1}{2} \left| -x(x-y) - y(y-(-x)) + (x+y)((-y)-x) \right| \]Simplify to get:\[= \frac{1}{2} \left| -2xy + y^2 - x^2 \right|\]
5Step 5: Express \\( z \\) in terms of Modulus
Given that the area is 600 square units, equate and solve:\[\frac{1}{2} \left| -2xy + y^2 - x^2 \right| = 600\]The formula in given the reason states: \( \frac{3}{2} |z^2| \), equating both gives:\[\frac{3}{2} |z^2| = 600 \]
6Step 6: Solve for \\( |z| \\)
From \[\frac{3}{2} |z^2| = 600\]solve for \(|z|\):\[|z^2| = 400 \]So, \(|z| = 20\).
7Step 7: Conclusion
This confirms the assertion that if the area of the triangle is 600 square units, then the modulus of \(|z|\) must be 20, as both calculations match.
Key Concepts
Argand PlaneTriangle AreaModulus
Argand Plane
The concept of the Argand Plane is fundamental when dealing with complex numbers. It is a two-dimensional plane where each complex number is represented as a point. The horizontal axis is the real part, while the vertical axis is the imaginary part of a complex number. Hence, a complex number like \( z = x + yi \) gets a unique spot on this plane with coordinates \((x, y)\). The use of the Argand Plane allows us to transform complex number operations into geometric interpretations, making them easier to visualize and understand. This is particularly handy when considering transformations, rotations, or reflections accomplished by multiplying complex numbers. In the exercise, it describes three complex numbers used as vertices on this plane to form a triangle. Understanding how each complex number translates into a coordinate on the plane is crucial for calculating geometric properties such as areas.
Triangle Area
Understanding how to calculate the area of a triangle using vertices in a plane is a critical skill in geometry and is applied here in the Argand Plane. Given three points \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), the area \(T\) of the triangle can be calculated by the formula:
- \[ T = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| \]
Modulus
The modulus of a complex number is analogous to the absolute value for real numbers. It represents the distance from the origin to the point \(z\) on the Argand Plane. For a complex number \(z = x + yi\), the modulus is calculated using:
- \[ |z| = \sqrt{x^2 + y^2} \]
Other exercises in this chapter
Problem 181
The roots of the equation \(z^{4}-z^{3}+z^{2}-z+1=0\) are \(\cos \left(\frac{p \pi}{5}\right)+i \sin \left(\frac{p \pi}{5}\right)\) where \(p=\) (A) \(1,3,5,7,9
View solution Problem 182
\hline Column-I \begin{tabular}{l} Column-II \\ \hline (A) \(i-1\) \end{tabular} (I) If \(z_{r}=\cos \left(\frac{\pi}{3^{r}}\right)+i \sin \left(\frac{\pi}{3^{r
View solution Problem 188
Assertion: If \(|z-1|+|z+3| \leq 8\), then the range of values of \(|z-4|\) is \([1,9]\) Reason: \(|z-1|+|z+3| \leq 8 \Rightarrow z\) lies inside or on the elli
View solution Problem 190
Assertion: The locus of the points representing the complex numbers satisfying \(|z|-2=0,|z-i|-\) \(|z+5 i|=0\) is the single point \((0,-2)\) Reason: If \(z\)
View solution