A) \[|z|\]
B) \[|z{{|}^{2}}\]
C) \[2|z{{|}^{2}}\]
D) \[\frac{1}{2}|z{{|}^{2}}\]
E) \[\frac{1}{2}|z|\]
Correct Answer: D
Solution :
\[iz=z{{e}^{i\pi /2}}\] This implies that 12 is the vector obtained by rotating vector z in anticlock wise direction through\[90{}^\circ \]. \[\therefore \] \[OA\bot AB\] So area of \[\Delta OAB=\frac{1}{2}OA\times OB\] \[=\frac{1}{2}|z||iz|=\frac{1}{2}|z{{|}^{2}}\]You need to login to perform this action.
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