The area of the triangle with vertices $z,iz\:\:and\:\:-iz$ is ?

Answer 1

Toolbox:

• If $z=x+iy$, then the point represented by z is $(x,y)$
• $i^2=-1$
• $|z|=\sqrt{x^2+y^2}$
• Area of the triangle with vertices $(x_1,y_1),\:(x_2,y_2)\:and\:(x_3,y_3)$ is given by $\large \frac{1}{2}$$\left|\begin {array}{ccc}x_1 & y_1 & 1\\x_2 & y_2 &1\\x_3 &y_3&1\end {array}\right|$

Let $z=x+iy$

$iz=i(x+iy)=xi+i^2y=-y+xi$

$-iz=-i(x+iy)=-xi-i^2y=y-xi$

Then the vertices of the triangle are given by

$A(z)=(x,y), \:\:B(iz)=(-y,x)\:\:and\:\:C(-z)=(y,-x)$

Area of $\Delta \:ABC=\large \frac{1}{2}$$\left|\begin {array}{ccc}x & y & 1\\-y & x &1\\y &-x&1\end {array}\right|$

$=\large\frac{1}{2}$$|1(xy-xy)-1(-x^2-y^2)+1(x^2+y^2)|$

$=x^2+y^2=|z|^2$