# Quod Erat Demonstrandum

## 2007/11/19

### [U] Solve sin(z) = 2

Filed under: University Mathematics — johnmayhk @ 3:21 下午
Tags:

A F.5 student, Yan, discussed with me something about “complex angles" in trigonometric formulae, in lunch hour, he asked me the way of solving

$\sin(\theta) = 2$

I just post it here. Starting with the Euler’s formula

$e^{i\theta} = \cos(\theta) + i\sin(\theta)$

We can extend the angle $\theta$ to complex number by using the above formula. Replace $\theta$ by $-\theta$, we have

$e^{-i\theta} = \cos(\theta) - i\sin(\theta)$

Subtracting, yields

$e^{i\theta} - e^{-i\theta} = 2i\sin(\theta)$, thus

$\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$

Now, putting $\sin(\theta) = 2$, we have

$\frac{e^{i\theta} - e^{-i\theta}}{2i} = 2$
$u^2 - 4iu - 1 = 0$ where $u = e^{i\theta}$

$u = \frac{4i \pm \sqrt{(4i)^2 - 4(1)(-1)}}{2} = \frac{4i \pm \sqrt{-16 + 4}}{2} = (2 \pm \sqrt{3})i$

Thus,

$e^{i\theta} = (2 \pm \sqrt{3})i$
$e^{i(x + iy)} = (2 \pm \sqrt{3})e^{(2n\pi + \pi/2)i}$

Here, we put $\theta$ as a complex number $x + iy$ and try to find out values of $x$ and $y$. Also, by putting $\theta = 2n\pi + \frac{\pi}{2}$ into the Euler’s formula, we have $i = e^{(2n\pi + \pi/2)i}$. Continue,

$e^{-y}e^{ix} = (2 \pm \sqrt{3})e^{(2n\pi + \pi/2)i}$

$\therefore x = 2n\pi + \pi/2$
$e^{-y} = 2 \pm \sqrt{3}$, taking logarithm, yields
$\therefore y = -\ln(2 \pm \sqrt{3})$

Last time, I’d derived the formula of $\ln(z)$ to Yan, hence it is done.