As

sinz = 1 2i (eiz - e-iz) = 2

then, substituting w=e^iz, we have

w - 1 w = 4i.

So w² - 4iw -1 = 0 and the solutions of this quadratic equation are:

w = eiz = 4i ±Ö- 12 2 = i(2±Ö3).

Taking logarithms gives

iz = loge i(2±Ö3) = loge i + loge(2±Ö3) = i p 2 + loge (2±Ö3).

Dividing by i gives the solutions

z =

p 2

-i loge (2±Ö3)

but since sinz is periodic with period 2p the set of all solutions is given by

z = p 2 - i log(2±Ö3) +2np.

[Editor's note:

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

The step in the solution above

loge i =

p 2

follows from the definition of the complex logarithm function using |i|=1 and

argi =

p 2

. The logarithms of z are the numbers l such that e^l = z, that is:

l = log|z| + i (argz + 2pn)

for integer n . It is easy to check:

elog|z| + i (argz + 2pn) = elog|z| ×eiarg z×e2pni = |z|eiargz = 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 2pi because e^2pni=1.]