If z, iz and z+iz are the vertices of a trian | vedclass

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(B) Let $z = x + iy$. Then $|z| = \sqrt{x^2 + y^2} = 4$,so $x^2 + y^2 = 16$.The vertices of the triangle are $A = z$,$B = iz$,and $C = z + iz$.Note th

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$8$

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(B) Let $z = x + iy$. Then $|z| = \sqrt{x^2 + y^2} = 4$,so $x^2 + y^2 = 16$.The vertices of the triangle are $A = z$,$B = iz$,and $C = z + iz$.Note that $C - A = iz$ and $C - B = z$.The vectors representing the sides $AC$ and $BC$ are $iz$ and $z$ respectively.The area of the triangle formed by complex numbers $z_1, z_2, z_3$ is given by $\frac{1}{2} |	ext{Im}(\bar{z_1}z_2 + \bar{z_2}z_3 + \bar{z_3}z_1)|$.Alternatively,since the triangle is formed by $z, iz$,and $z+iz$,it is a right-angled triangle with vertices at the origin $O(0,0)$,$A(z)$,and $B(iz)$ if we consider the triangle $OAB$ where $O, z, z+iz$ form a triangle with base $|z|$ and height $|iz|$.Actually,the vertices are $z, iz, z+iz$. This is a right-angled triangle because the vector $z$ and $iz$ are perpendicular.The lengths of the two legs are $|z|$ and $|iz|$.Since $|iz| = |i||z| = |z| = 4$,the area is $\frac{1}{2} 	imes |z| 	imes |iz| = \frac{1}{2} 	imes 4 	imes 4 = 8$ sq. units.

If $z, iz$ and $z+iz$ are the vertices of a triangle and if $|z|=4$,then the area (in sq. units) of that triangle is:

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