Quod Erat Demonstrandum

2007/11/19

[U] Solve sin(z) = 2

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

By quadratic formula,

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.

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