First Post

Consider the equation

\left(z+2\right)^{12}=z^{12}\quad (*)

which cropped up in the (somewhat more than averagely difficult) complex numbers question (Q10) of the 2005 Further Pure Maths exam.

How many roots does it have in the field of complex numbers? The answer is 11, not 12; nothing more than a little trick question. The coefficient of
z^{12} is zero.

How many of the 11 roots are not real? Easy to check - there are 10 - complex conjugate pairs, of course, but more interesting is that they are linked directly to the 12th roots of unity.

\left(\frac{z+2}{z}\right)^{12}=1

Thus each non-real twelfth root of unity corresponds via a Mobius transformation to one root of (*).

The question asks the student to demonstrate that each of these roots takes the form

-1 -i\cot{\left(\frac{k\pi}{12}\right),\quad k=\pm 1, \pm 2, \pm 3, \pm 4, \pm 5

Those students who successfully solve the question generally replace z with \cos{\theta}+i\sin{\theta}, and follow the standard procedure of multiplying numerator and denominator by the complex conjugate of the denominator. This requires several lines of algebra and some deftness.

The student is then asked to jump through a few hoops and play a few trigonometric and series tricks to derive the beautiful but slightly crazy formula:

\sin^{2}{\left(\frac{\pi}{12}\right)}\sin^{2}{\left(\frac{2\pi}{12}\right)}\sin^{2}{\left(\frac{3\pi}{12}\right)}\sin^{2}{\left(\frac{4\pi}{12}\right)}\sin^{2}{\left(\frac{5\pi}{12}\right)}=\frac{512}{3}

But hang on a minute! I really felt that there ought to be more to this question than just mashing the equation using a few key complex number properties.

A more elegant way to understand the equation is to write it like this:

\left(z+2\right)^{12}=z^{12} \quad \Rightarrow \quad (z+2)= \omega z

where \omega can be any of the twelfth roots of unity. Now, since all roots of unity have modulus 1, it follows that

|z+2|=|z|

for every solution of (*). This means that all the solutions lie on the perpendicular bisector of the line segment between the points -2 and 0 on the real line, i.e. all solutions have

\textrm{Re}(z)=-1

and it remains to show that for each twelfth root of unity there is precisely one point on that perpendicular bisector with an imaginary part of

\cot{\left(\frac{\arg{(\omega)}}{2}\right)}

(ignoring sign), which can be easily proved geometrically by considering the isosceles triangle enclosed by the directed line segments between (-2), (0) and (z).

It is also interesting to see why according to this geometric method there are only 11 solutions and not 12; the 12th solution corresponds to the root of unity \omega = +1, which has argument zero. Here \cot{\left(\frac{\arg{(\omega)}}{2}\right)} = \pm \infty (of course, more accurately, I should say "undefined" rather than infinite), which reflects the fact that, in some sense, the point at infinity is equidistant from -2 and 0!

These kinds of problems have a wonderful richness in that they can be attacked purely geometrically, purely algebraically or as a combination of the two.