The image of the line $\operatorname{Re}(z) = 1$ under $f(z) = (1+i)\frac1z - 2i$

Question

I'm investigating the image of the line $\operatorname{Re}(z) = 1$ under $f(z) = (1+i)\frac1z - 2i$

Since the above is a linear fractional transformation, I know that this line must either map to a line or a circle. So I select three points on the line: $1$, $1+i$, and $1-i$.

However, under the equation above, I get that:

$f(1) = 1 - i$

$f(1+i) = 1 - 2i$

$f(1-i) = -i$

Quite clearly, this isn't a line, nor a circle. So I was wondering what the image under this transformation was?

Answer 1

So $\Re(z) = 1 \Rightarrow |z| = |z-2|$, then we have $$w :=f(z) = (1+i)\frac{1}{z} - 2i$$Making $z$ the subject, we have $$z = \frac{1+i}{w+2i}$$ Now, let's substitute this into the locus for $z$ to see the locus of $w$: $$\left |\frac{1+i}{w+2i}\right| = \left| \frac{1+i}{w+2i} - 2\right|$$ $$\Rightarrow \frac{|1+i|}{|w+2i|} = \frac{|1+i-2w-4i|}{|w+2i|}$$ $$\Rightarrow \sqrt{2} = |1-3i - 2w|$$ Let $w = u +iv$, $$\Rightarrow \sqrt{2} = |1- 3i - 2u - 2iv|$$ $$\Rightarrow \sqrt{2} = |(1-2u) + (-2v-3)i|$$ $$\Rightarrow 2 = (1+2u)^2 + (2v+3)^2$$ $$\Rightarrow \left(u+\frac{1}{2}\right)^2 + \left(v+ \frac{3}{2} \right)^2 = 8$$ So points on the real axis in the $z$-plane are mapped to points on the circle defined by the equation above in the $w$-plane.

Answer 2

$f(z) = \frac 1z$ will map a line that does not go through the origin to circle where 0 lies on the circle, and vice versa.

$f(z) = (1+i)\frac 1z$ adds a rotation and dilation factor. $f(z) = \frac {(1+i)}{z} - 2i$ adds a translation.

$\{z:\Re(z) = 1\}$ will map to a circle.

Answer 3

Your function is a so-called Möbius transformation: $$f(z) = (1+i)\frac1z - 2i = \frac{-2iz+(1+i)}{z}= \frac{az+b}{cz+d}\;\; (ad-bc \neq 0)$$ It maps generalized circles (lines and circles) onto generalized circles. $Re(z)= 1$ is a generalized circle (through $\infty$). So, the image of $Re(z)$ under $f$ is the circle through $$f(\infty) = -2i,\, f(1) = 1-i, \, f(1+i) = 1-2i$$