You may also like 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

Age 16 to 18 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 $e^{2\pi ni}=1$.