Question

Area of the triangle formed by the complex numbers \(z\), \(iz\) and \(z+iz\) in the Argand diagram as vertices is:

• A. \(\frac{1}{2}|z|^2\)
• B. \(\frac{1}{2}z^2\)
• C. \(z^2\)
• D. \(|z|^2\)

Hint:

For area of triangle in Argand diagram, use \(\text{Area} = \frac{1}{2}|\text{Im}\{(z_2-z_1)\overline{(z_3-z_1)}\}|\).

Answer 1

The Correct Option is A

Step 1: Recognize the parallelogram formed by 0, z, iz, z+iz.
Since \(z+iz\) is just \(z\) shifted by the vector \(iz\), the four points \(0,z,z+iz,iz\) form a parallelogram with adjacent sides given by the vectors \(z\) and \(iz\).

Step 2: Split the parallelogram along its diagonal.
The diagonal joining \(z\) and \(iz\) splits the parallelogram into two congruent triangles, one of which is exactly the triangle with vertices \(z, iz, z+iz\) asked about here.
So its area is half the parallelogram's area.

Step 3: Compute the parallelogram area from the two side vectors.
For vectors \(z\) and \(iz\), the parallelogram area is \(|\operatorname{Im}(\overline{z}\cdot iz)|\).
\(\overline{z}\cdot iz = i\,z\overline{z} = i|z|^2\), which is purely imaginary with imaginary part \(|z|^2\).
So the parallelogram has area \(|z|^2\), and the triangle has half of that.

Step 4: Conclusion.
\[ \boxed{\frac{1}{2}|z|^2} \]