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Question: If \(z = x + iy\), then area of the triangle whose vertices are points z, iz and \(z + iz\) is...

If z=x+iyz = x + iy, then area of the triangle whose vertices are points z, iz and z+izz + iz is

A

2∣z∣22|z|^{2}

B

12∣z∣2\frac{1}{2}|z|^{2}

C

∣z∣2|z|^{2}

D

32∣z∣2\frac{3}{2}|z|^{2}

Answer

12∣z∣2\frac{1}{2}|z|^{2}

Explanation

Solution

Sol. Let z=x+iyz = x + iy , z+iz=(x−y)+i(x+y)z + iz = (x - y) + i(x + y) and iz=−y+ixiz = - y + ix

If A denotes the area of the triangle formed by z,z+izz,z + iz and iziz, then

$A = \frac{1}{2}\left| \begin{matrix} x \ x - y \

  • y \end{matrix}\begin{matrix} y \ x + y \ x \end{matrix}\begin{matrix} 1 \ 1 \ 1 \end{matrix} \right|$

Applying transformation R2→R2−R1−R3R_{2} \rightarrow 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}$.