The area of the triangle whose vertices are represented by the complex numbers 0,z,zeiα,(0<α<π) equals:
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a
12|z|2cosα
b
12|z|2sinα
c
12|z|2sinαcosα
d
12|z|2
answer is B.
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Detailed Solution
Vertices are 0=0+i,0,z=x+iy and zeiα=(x+iy)(cosα+isinα) =(xcosα−ysinα)+i(ycosα+xsinα) Area =12|001xy1(xcosα−ysinα)(ycosα+xsinα)1| =12[xycosα+x2sinα−xycosα+y2sinα] =12sinα(x2+y2)=12|z|2sinα [∵ |z|=(x2+y2)]