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

  • 3
    $\begingroup$ It is not a half-turn around the origin. It's a half-turn around $1$. $\endgroup$ Jan 30 at 18:34
  • $\begingroup$ @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. $\endgroup$ Jan 30 at 19:13

2 Answers 2


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.

  • $\begingroup$ 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 ! :-) $\endgroup$ Jan 30 at 19:32

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