# Thread: Triangle under 1/z in the complex field...

1. ## Triangle under 1/z in the complex field...

Hey guys.

I'm trying to find the "picture" (sorry if it's not the word) of the triangle under 1/z in the complex field.
I think that [0,2i] -> [-i/2, inf], [0,2] -> [-1/2, inf] but I'm not sure about the 3rd side of the triangle. I tried putting in some number, but all it did is to confuse me even more.
Can I please have some help?

Thanks a lot.

2. Originally Posted by asi123
Hey guys.

I'm trying to find the "picture" (sorry if it's not the word) of the triangle under 1/z in the complex field.
I think that [0,2i] -> [-i/2, inf], [0,2] -> [-1/2, inf] but I'm not sure about the 3rd side of the triangle.
If I understand the question correctly, you have a triangle in the complex plane, with vertices at 0, 2 and 2i, and you want to find its image under the map $z\mapsto 1/z$.

The image of the side [0,2] is the segment $[\frac12,\infty)$ on the real axis. The image of the side [0,2i] is the segment $[-\frac12i,-\infty i)$ on the imaginary axis.

The remaining side is a segment of the line $x+y=2$, where $z = x+iy$. Let $w=1/z$ and write $w = u+iv$. Then $x+iy = \dfrac1{u+iv} = \dfrac{u-iv}{u^2+v^2}$. Take real and imaginary parts to see that $x = u/(u^2+v^2)$ and $y = -v/(u^2+v^2)$. Then the equation $x+y=2$ becomes $u-v = 2(u^2+v^2)$. Rewrite this as $\bigl(u-\frac14\bigr)^2 + \bigl(v+\frac14\bigr)^2 = \frac18$. This represents a circle centred at $\frac14-\frac14i$, with radius $1/\sqrt8$.

The image of the third side of the triangle is therefore part of that circle, namely the part lying in the fourth quadrant. The image of the triangle and its interior is the whole of the fourth quadrant apart from the interior of that circle.