Question

If $z$, $iz$, and $z+iz$ are the vertices of a triangle whose area is $2$ units then the value of $\left|z\right|$is

• A. $4$
• B. $2$
• C. $-2$
• D. None of these

Answer 1

The correct option is B

$2$

Explanation for the correct option.

Step 1. Find the vertices of the triangle.

Let $z=x+iy$ be the complex number, now $iz$ is defined as:

$\begin{array}{rcl}iz& =& i(x+iy)\\ & =& ix+i^{2}y\\ & =& ix+\left(-1\right)y\left[i^2=-1\right]\\ & =& -y+ix\end{array}$

And the complex number $z+iz$ is defined as:

$\begin{array}{rcl}z+iz& =& x+iy+\left(-y+ix\right)\\ & =& \left(x-y\right)+i\left(x+y\right)\end{array}$

As $z$, $iz$, and $z+iz$ are the vertices of a triangle, so the coordinates of the vertices are: $\left(x,y\right),\left(-y,x\right),\left(\left(x-y\right),\left(x+y\right)\right)$.

Step 2. Find the value of $\left|z\right|$

The area of the triangle having vertices as $\left(x,y\right),\left(-y,x\right),\left(\left(x-y\right),\left(x+y\right)\right)$ is given as:

$\begin{array}{rcl}\mathrm{Area}& =& \frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ -y& x& 1\\ x-y& x+y& 1\end{array}\right|\\ & =& \frac{1}{2}\left|\left[x\left(x-\left(x+y\right)\right)-y\left(-y-\left(x-y\right)\right)+1\left(-y\left(x+y\right)-x\left(x-y\right)\right)\right]\right|\\ & =& \frac{1}{2}\left|\left[x\left(x-x-y\right)-y(-y-x+y)+\left(-xy-y^2-x^2+xy\right)\right]\right|\\ & =& \frac{1}{2}\left|-xy+xy-y^2-x^2\right|\\ & =& \frac{1}{2}\left|-\left(x^2+y^2\right)\right|\\ & =& \frac{1}{2}\left(x^2+y^2\right)\end{array}$

But the area of the triangle is $2$ units. So,

$$\begin{gathered} \frac{1}{2}\left(x^2+y^2\right)=2 \\ \Rightarrow x^2+y^2=4 \end{gathered}$$

Now, for the complex number $z=x+iy$, the modulus is given as:

$\begin{array}{rcl}\left|z\right|& =& \sqrt{x^2+y^2}\\ & =& \sqrt{4}\left[x^2+y^2=4\right]\\ & =& 2\end{array}$

So, the value of $\left|z\right|$ is $2$.

Hence, the correct option is B.