### Roots and Coefficients

If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?

### Target Six

Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.

### 8 Methods for Three by One

This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best?

# Complex Sine

##### Stage: 5 Challenge Level:
Thank you to Barinder Singh Banwait, Langley Grammar School; Angus Balkham from Bexhill College and Derek Wan for your excellent solutions.

As $$\sin z = {1\over 2i}\left(e^{iz} - e^{-iz}\right) = 2$$ then, substituting $w=e^{iz}$, we have $$w - {1\over w} = 4i.$$ So $w^2 - 4iw -1 = 0$ and the solutions of this quadratic equation are: $$w = e^{iz} = {4i \pm \sqrt{- 12} \over 2} = i\left(2\pm \sqrt 3\right).$$ Taking logarithms gives $$iz = \log_e i\left(2\pm \sqrt 3\right) = \log_e i + \log_e\left(2\pm \sqrt 3\right) = i{\pi\over 2} + \log_e \left(2\pm \sqrt 3\right).$$ Dividing by $i$ gives the solutions $z = {\pi \over 2} -i \log_e \left(2\pm \sqrt 3\right)$ but since $\sin z$ is periodic with period $2\pi$ the set of all solutions is given by $$z ={\pi \over 2} - i \log \left(2\pm\sqrt 3\right) +2n\pi.$$

The same method works to give solutions for $\sin z = a$ where $a$ is any complex number.

The step in the solution above $\log_e i = {\pi \over 2}$ follows from the definition of the complex logarithm function using $|i|=1$ and $\arg i = {\pi \over 2}$. The logarithms of $z$ are the numbers $\lambda$ such that $e^{\lambda} = z$, that is: $$\lambda = \log |z| + i (\arg z + 2\pi n)$$ for integer $n$ . It is easy to check: $$e^{\log |z| + i (\arg z + 2\pi n)}= e^{\log |z|} \times e^{i\arg z}\times e^{2\pi ni} = |z|e^{i\arg z} = z$$ giving the complex number $z$ in modulus-argument form. Every complex number has infinitely many complex logarithms each differing by an integer multiple of $2\pi i$ because $e^{2\pi ni}=1$.