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Need to find an isometry which would rotate about the point $i$ by $\frac{\pi}{4}$.

So I was thinking that first I return the given point to orign, make the rotation and then translate back, right?

That would look like:

$$f(z)=z\frac{1+i}{\sqrt2}-\bigg(\frac{1+i}{\sqrt2}\bigg)+i$$

Is this correct?

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2 Answers 2

up vote 0 down vote accepted

Your answer is almost correct, but not quite: $$f(z) = (z-i)\frac{1+i}{\sqrt{2}}+i = z\frac{1+i}{\sqrt{2}}-\left(\frac{-1+i}{\sqrt{2}}\right)+i.$$

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First translate $i$ to the origin $$f_1(z):=z-i$$ Then rotate $$f_2(z):=e^{\frac{2\pi i}{4}}z=\frac{1+i}{\sqrt{2}}z$$ Finally do the opposite translation $$f_3(z):=z+i$$ Composing you get $$f:=f_3\circ f_2\circ f_1:z\longmapsto \frac{1+i}{\sqrt{2}}(z-i)+i$$

You can check that your $f$ doesn't work noticing that it doesn't fix $i$.

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