Question

Area of the triangle on the Argand plane formed by the complex numbers z, iz and z +iz is

Answer 1

Let $A=z,B=izandC=z+iz$

Let $z=x+iy$ then the values of $|A-B|,|B-C|$ and $|C-A|$ forms an isosceles triangle with $AC=BC$.

It is a property of an isosceles triangle that if we join $C$ and mid point of $A$ and $B$ (let mid point be $P$), it will be perpendicular to $AB$.

$AREA=\frac{1}{2}\times AB\times PC$

$P=$ mid point of $A$ and $B=\frac{A+B}{2}=\frac{z+iz}{2}$

Now $PC=|z+iz-\frac{z+iz}{2}|=\left|\frac{z+iz}{2}\right|$

$AB=|z-iz|$

Therefore area $=\frac{1}{2}\times \frac{|z+iz|}{2}\times |z-iz|=\frac{|z^2+z^2|}{4}=\frac{|z{|}^2}{2}$