# Put a transformation under the form of a rotation in the complex plane

On the complex plane, I have a transformation "T" such that :

$z' = (m+i)z + m - 1 - i$ ($z'$ is the image and $z$ the preimage, $z$ and $z'$ are both complex number)

and $m$ is a real number.

I'd need to determine "$m$" such that this transformation "T" is a rotation.

I know a rotation can be written under the form : $z'- w = k (z - w)$ with "$w$" the complex number associated with the center and "$k$" a complex number modulus 1. But I can't find how to put "T" under the form of a rotation.

Some hint would be very appreciated, Thanks.

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@Ayman Hourieh : thanks for the edit :). – Jecimi Jun 3 '12 at 21:34

Rotations are described by formula $z'-w=k(z-w)$, i.e. $z'=kz+w(1-k)$.

You are given transformation "T" of the form $z'=(m+i)z+m-1-i$. Hence $$k=m+i$$ Using condition $|k|=1$ you can determine $m$.

The rest is clear.

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Thanks a lot for your help ! – Jecimi Jun 3 '12 at 21:22

Use the form you have to get $z'=kz-kw+w=kz+(1-k)w$. Comapre it to the form you have. You know that $k=m+i$ and that $(1-k)w=m-1-i$. Since $k$ has to have modulus $1$, you can find $k$. From there find $w$.

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Thank you for answering, it helped me a lot. – Jecimi Jun 3 '12 at 21:22