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Find the area of triangle formed by $-z, iz$ and $z-iz$.

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    It will be $|z|^2$ times the area of $-1, i, 1-i$ or $0, i+1, 2-i$. You could use determinants. – 2017-02-04
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    Have you drawn this triangle for a few values of $z$? What kind of triangle is it? How do you find the area of such a triangle? Please, tell us a bit more, not about the problem, but what _you_ have done in an attempt to solve it. – 2017-02-04
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    $z$ can be a complex number, right? – 2017-02-05

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You can add to all of the complex numbers $z$, that doesn't change the area. Then we have $A=0,B=z+iz,C=2z-iz$. Use the representation $z=x+iy$. We look at the coordinates in the complex plane of $B$ and $C$:

$$B=(x-y,y+x)^T, \ \ C=(2x+y,2y-x)^T$$

Then the area is given by $V = \frac{1}{2}|\det(BC)|$, since the absolute value of the determinant of a $2 \times 2$ matrix is the area of the parallelogram created by the two collumn vectors. A simple calculation shows that $|\det(BC)| = 3(x^2+y^2)=3|z|^2$. Hence $V=\frac{3}{2}|z|^2$.