Question

For the following question, enter the correct numerical value upto TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numeric value 5 will be written as 5.00 and 2.346 will be written as 2.35) If \( z \), \( iz \) and \( z+iz \) are the vertices of a triangle and if \( |z| = 4 \), then the area (in sq. units) of that triangle is ______.

Hint:

When complex numbers represent points on the Argand plane, convert them into coordinates and use the determinant method to find the area of the triangle.

Answer 1

Correct Answer: 8

Step 1: Let \[ z = a + ib \] Then, \[ iz = i(a+ib) = -b + ia \] and \[ z + iz = (a-b) + i(a+b) \] Thus, the vertices of the triangle are: \[ (a,b),\; (-b,a),\; (a-b,a+b) \]
Step 2: Form two vectors with origin at \( z \): \[ \vec{v}_1 = iz - z = (-b-a,\; a-b) \] \[ \vec{v}_2 = (z+iz) - z = (-b,\; a) \]
Step 3: Area of triangle formed by vectors \( \vec{v}_1 \) and \( \vec{v}_2 \) is: \[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} -b-a & a-b \\ -b & a \end{vmatrix} \right| \]
Step 4: Evaluate the determinant: \[ (-b-a)(a) - (a-b)(-b) = -ab - a^2 + ab - b^2 = -(a^2 + b^2) \]
Step 5: \[ \text{Area} = \frac{1}{2}(a^2 + b^2) \] Given: \[ |z| = \sqrt{a^2 + b^2} = 4 \Rightarrow a^2 + b^2 = 16 \]
Step 6: \[ \text{Area} = \frac{1}{2} \times 16 = 8 \] Final Answer (up to two decimal places): \[ \boxed{8.00} \]