Question

If $z=x+iy$, then the area of the triangle whose vertices are points $z$, $iz$ and $z+iz$ is
A. $2{{\left| z \right|}^{2}}$
B. $\dfrac{1}{2}{{\left| z \right|}^{2}}$
C. ${{\left| z \right|}^{2}}$
D. $\dfrac{3}{2}{{\left| z \right|}^{2}}$

Answer 1

Hint: In this question, we are to find the area of the given triangle. By applying a matrix determinant for the given vertices, the required area is obtained.

Formula used: The complex number $(x,y)$ is represented by $x+iy$.
If $z=x+iy\in C$, then $x$ is called the real part and $y$ is called the imaginary part of $z$. These are represented by $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ respectively.
$z=x+iy$ be a complex number such that $\left| z \right|=r$ and $\theta $ be the amplitude of $z$. So, $\cos \theta =\dfrac{x}{r},\sin \theta =\dfrac{b}{r}$
And we can write the magnitude as
 $\begin{align}
  & \left| z \right|=\left| x+iy \right| \\
 & \Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
\end{align}$
Thus, we can write
$z=x+iy=r\cos \theta +ir\sin \theta =r(\cos \theta +i\sin \theta )$
This is said to be the mod amplitude form or the polar form of $z$.
Where $\cos \theta +i\sin \theta $ is denoted by $cis\theta $ and the Euler’s formula is $\cos \theta +i\sin \theta ={{e}^{i\theta }}$

Complete step by step solution: Given that, $z=x+iy$.
The vertices of the triangle are $z$, $iz$ and $z+iz$
So, we can write
$z=x+iy$
$\begin{align}
  & iz=i(x+iy) \\
 & \text{ }=ix+{{i}^{2}}y \\
 & \text{ }=-y+ix \\
\end{align}$
$\begin{align}
  & z+iz=x+iy-y+ix \\
 & \text{ }=(x-y)+i(x+y) \\
\end{align}$
Then, the area of the given triangle is
$A=\dfrac{1}{2}\left| \begin{matrix}
   x & y & 1 \\
   x-y & x+y & 1 \\
   -y & x & 1 \\
\end{matrix} \right|$
To find the determinant, transformations on rows of the matrix should be applied.
The required transformations are ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}-{{R}_{3}}$
Then, we get
$\begin{align}
  & A=\dfrac{1}{2}\left| \begin{matrix}
   x & y & 1 \\
   x-y & x+y & 1 \\
   -y & x & 1 \\
\end{matrix} \right| \\
 & \text{ }=\dfrac{1}{2}\left| \begin{matrix}
   x & y & 1 \\
   0 & 0 & -1 \\
   -y & x & 1 \\
\end{matrix} \right| \\
 & \text{ }=\dfrac{1}{2}\left[ x(x)-y(-y)+0 \right] \\
 & \text{ }=\dfrac{1}{2}\left( {{x}^{2}}+{{y}^{2}} \right) \\
 & \text{ }=\dfrac{1}{2}{{\left| z \right|}^{2}} \\
\end{align}$

Thus, Option (B) is correct.

Note: Here, we may use transformations, or directly we can calculate the determinant of the matrix for finding the area of the triangle. Here we used the matrix method because the given vertices are complex numbers. By applying appropriate operation, the required area is obtained.