# Transformations of the Argand plane described in geometrical terms

So far I've learned that the transformation of $$\;f(z)=\overline z\;$$ is the same as a reflection in the x-axis the transformation of $$f(z)=iz$$ is the same as an anti-clockwise rotation of 90 deg about the origin and I'm quite happy with how to do those.

However a question in my book is "describe the transformation of $$f(z)=2-z"$$. The given answer is that it is a half turn rotation about the origin and I just cannot work out where this comes from.

Any guidance would be much appreciated. Thank you

• It is not a half-turn around the origin. It's a half-turn around $1$. Jan 30 at 18:34
• @JoséCarlosSantos The problem is that the OP (i.e. original poster) needs a systematic way of reaching your conclusion, for all problems of this type. Personally, I regard the problem composer's idea of trying to combine a rotation + shift into a shifted rotation bad. The idea behind my answer is that it gives the OP a systematic method of attack. Jan 30 at 19:13

The transformation maps $$\,z \,\mapsto\, z'=2-z\,$$, which can be also written as $$\,\frac{1}{2}(z+z')=1\,$$. In other words, the midpoint of the segment between $$\,z\,$$ and $$\,z'\,$$ is the fixed point $$\,z_0=1\,$$, and therefore points $$\,z,z'\,$$ are symmetric about $$\,z_0\,$$, so the transformation is the point reflection across $$\,z_0=1\,$$.

Like all point reflections, the transformation can also be described as a rotation of angle $$\,\pi\,$$ about the central point, as OP's book appears to do.

Break it up into $$2$$ operations.

First, you are multiplying $$z$$ by $$-1$$.

So, if $$z = re^{(i\theta)}$$, and you multiply it by $$e^{(i\pi)}$$, the product is $$re^{i(\theta + \pi)}$$, which is a $$180^\circ$$ counter-clockwise rotation.

Then, you are adding $$2$$ to the product, which is a shift of $$2$$ units to the right.

So, the combined operations are:

• first, a rotation of $$180^\circ$$.

• then, a shift of $$2$$ units to the right, which is in the direction of the positive real numbers.

• Thank you both for your replies. I hadn’t worked out that the rotation was around 1 but I had gone by the method of multiplying z by -1 then moving 2 places to the right. A relief to know that there was a typo in the book and it’s not me ! :-) Jan 30 at 19:32