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

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2 \sqrt{2}

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Solution

The correct option is B 2
Let $z = x + i y \equiv (x, y)$
Then $i z = i x - y \equiv (- y, x), z + i z = x + i y + i x - y \equiv (x - y, x + y)$
Area $= \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} x & y & 1 - y & x & 1 x - y & x + y & 1 \end{matrix} ∣ ∣ ∣ ∣ = 2$
$\Rightarrow x^{2} + y^{2} = | z |^{2} = 4 \Rightarrow | z | = 2$

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