A) \[\frac{1}{2}|z{{|}^{2}}\cos \alpha \]
B) \[\frac{1}{2}|z{{|}^{2}}\sin \alpha \]
C) \[\frac{1}{2}|z{{|}^{2}}\sin \alpha \cos \alpha \]
D) \[\frac{1}{2}|z{{|}^{2}}\]
Correct Answer: B
Solution :
Vertices are \[0=0+i0,\] \[z=x+iy\] and \[z{{e}^{i\alpha }}=(x+iy)\,\,(\cos \alpha +i\sin \alpha )\] \[=(x\cos \alpha -y\sin \alpha )+i(y\cos \alpha +x\sin \alpha )\] \[\therefore \] Area \[=\frac{1}{2}\,\left| \,\begin{matrix} 0 & 0 & 1 \\ x & y & 1 \\ (x\cos \alpha -y\sin \alpha )\,\,\, & (y\cos \alpha +x\sin \alpha )\,\, & 1 \\ \end{matrix}\, \right|\] \[=\frac{1}{2}[xy\cos \alpha +{{x}^{2}}\sin \alpha -xy\cos \alpha +{{y}^{2}}\sin \alpha ]\] \[=\frac{1}{2}\sin \alpha ({{x}^{2}}+{{y}^{2}})\]\[=\frac{1}{2}|z{{|}^{2}}\sin \alpha \]\[[\because \,\,|z|=\sqrt{{{x}^{2}}+{{y}^{2}}}]\].
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