A) \[2|z{{|}^{2}}\]
B) \[\frac{1}{2}|z{{|}^{2}}\]
C) \[|z{{|}^{2}}\]
D) \[\frac{3}{2}|z{{|}^{2}}\]
Correct Answer: B
Solution :
Let\[z=x+iy\]; \[z+iz=(x-y)+i(x+y)\]and\[iz=-y+ix\] If A denotes the area of the triangle formed by \[z,z+iz\] and \[iz\], then \[A=\frac{1}{2}\left| \begin{matrix} x & y & 1 \\ x-y & x+y & 1 \\ -y & x & 1 \\ \end{matrix} \right|\] Applying transformation\[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}-{{R}_{3}}\], we get \[A=\frac{1}{2}\left| \begin{matrix} x & y & 1 \\ 0 & 0 & -1 \\ -y & x & 0 \\ \end{matrix} \right|=\frac{1}{2}({{x}^{2}}+{{y}^{2}})=\frac{1}{2}|z{{|}^{2}}\]
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