question_answer

If \[z=x+iy,\]then area of the triangle whose vertices are points \[z,\text{ }iz\]and \[z+iz\]is

A) \[2|z{{|}^{2}}\]               

B)   \[\frac{1}{2}|z{{|}^{2}}\]

C) \[|z{{|}^{2}}\]                 

D)   \[\frac{3}{2}|z{{|}^{2}}\]

Correct Answer: B

Solution :

\[iz=z{{e}^{i\pi /2}}\] This implies that \[iz\]is the vector obtained by rotating vector z in anticlockwise direction through \[90{}^\circ \] Area of \[\Delta OAB=\frac{1}{2}(OA)(AB)=\frac{1}{2}|z||iz|=\frac{1}{2}|z{{|}^{2}}\].