# Area of triangle formed by three complex numbers $-z,iz$ and $z-iz$ in the complex plane

The question bothering me is "Find the area of a triangle formed by the complex numbers $-z,iz,z-iz$ in the argand plane.

I'm really sorry, but I could not understand how to proceed. However to that extent I thought I start by assuming $z=a +ib$. Then $-z = -a-ib$ and $iz= -b+ia$. How do I proceed further?

• Here is meta.math.stackexchange.com/questions/5020/… for displaying numbers and functions. – Arbuja Jan 18 '16 at 12:01
• Suggestion: Make a sketch of the triangle for the case $z = 1$, and calculate its area. Then use geometry to figure out how area changes when all three vertices are multiplied by $z = r e^{i\theta}$. – Andrew D. Hwang Jan 18 '16 at 14:16

As you suggest, let $$z=a+ib\Rightarrow iz=-b+ia, z-iz=a+b+i(b-a)$$

Then the area of the triangle formed by the vertices $(-a,-b),(-b,a),(a+b,b-a)$ is $$\frac 12\left|\begin{matrix}-a&-b&a+b&-a\\-b&a&b-a&-b\end{matrix}\right|$$ $$=\frac 12\left|-a^2-b^2-b^2+ab-a^2-ab-ab-b^2+ab-a^2\right|$$ $$=\frac 32(a^2+b^2)$$

• $$\ldots = \frac 32 |z|^2$$ – CiaPan Jan 18 '16 at 16:26
• yes, of course! thanks – David Quinn Jan 18 '16 at 16:30
• thank you errally appreciate your help – Nitro phenol Jan 19 '16 at 9:29

For any complex number $z$, when you multiply it by $i$ you are rotating that vector by $90^{\circ}$ in the counterclockwise direction. So now try to visualize what kind of a triangle you will get in your question.

When $z=1$ you have the triangle $T$ with vertices $-1$, $i$, and $1-i$. Its area is ${3\over2}$. Multiplying these three vertices with the complex number $z$ is a similarity map, centered at $O$: The triangle $T$ is stretched linearly by the factor $|z|$ and rotated by ${\rm arg}(z)$. It follows that in the end you have a triangle of area ${3\over2}|z|^2$.

You can notice that all vertices are defined as $z$ multiplied by some constant: \begin{align} -z & = z \cdot (-1)\\ iz & = z \cdot i\\ z-iz &= z \cdot (1-i) \end{align} So you can analyze the triangle with vertices $-1, i, (1-i)$ (complex), that is $(-1,0), (0,1), (1,-1)$ (cartesian) first. You will find it is isosceles with a base of $\sqrt 2$ and a height $\frac 32 \sqrt 2$, so its area is $\frac 32$.

When you multiply all vertices by arbitrary $z$, they all get scaled with respect to $(0,0)$ by factor $|z|$ and rotated by angle $\arg z$. Rotation does not affect the area of a figure, but scaling applies to the area with a square, hence the answer $$\frac 32 \cdot |z|^2$$