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Find Z for sin(z)=4

The equation \(\sin(x) = 4\) does not have a solution because the range of the sine function is \([-1, 1]\). Therefore, it is not possible for \(sin(x)\) to equal 4, which is outside the range of the function.

But everything changed, when we start to use complex numbers! If \(x\) is a complex number, then the equation \(\sin(x) = 4\) has infinitely many solutions, because the sine function is periodic with a period of \(2\pi\).

To solve this problem, let's write sine in exponential form and for doing this let's use the following Tailor series and then make a sum of \(\cos(z) +i\sin(z)\):

\[ \sin(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{2n+1} = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots \\ \cos(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} z^{2n} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \cdots \\ e^{iz} = \sum_{n=0}^{\infty} \frac{(iz)^n}{n!} = 1 + iz - \frac{z^2}{2!} - i\frac{z^3}{3!} + \frac{z^4}{4!} + i\frac{z^5}{5!} - \cdots \\ \cos(z) = i\sin(z) = 1 + iz - \frac{z^2}{2!} - i\frac{z^3}{3!} + \frac{z^4}{4!} + i\frac{z^5}{5!} - \cdots = e^{iz}\]

now let's calculate \(e^{-iz}\):

\[e^{-iz} = \cos(-z) + i\sin(-z) = cos(z) - i\sin(z)\]

and now we can write that

\[e^{iz} - e^{-iz} = 2i\sin(-z)\]

\[\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}\]

Now, let's go back to our original problem and use the values given.

\[ 4 = \frac{e^{iz} - e^{-iz}}{2i}; z = ? \\ e^{iz} - e^{-iz} = 8i \\ (e^{iz} - e^{-iz})\times e^{iz} = (8i)\times e^{iz} \\ (e^{iz})^2 - 1 = (8i)\times e^{iz} \\ y = e^{iz} \\ y^2 - 8iy - 1 = 0 \\ y = \frac{8i \pm \sqrt{(-8i)^2 - 4\times(-1)}}{2} = \frac{8i \pm \sqrt{-64 + 4}}{2} = \frac{8i \pm 2i\sqrt{15}}{2} = i(4 \pm \sqrt{15}) \\ y = i(4 \pm \sqrt{15}) \\ e^{iz} = i(4 \pm \sqrt{15}) \\ \ln(e^{iz}) = \ln(i(4 \pm \sqrt{15})) \\ iz = \ln(i(4 \pm \sqrt{15})) = \ln(i) + \ln((4 \pm \sqrt{15})) \\ z = \frac{1}{i}\times\ln(i) + \frac{1}{i}\times\ln((4 \pm \sqrt{15})) \\ \ln(i) = i\times(\frac{\pi}{2} + 2\pi n) \\\]

And now we can combine everything together:

\[ \sin(z) = 4 => \\ z = (\frac{\pi}{2} + 2\pi n) - i\ln((4 \pm \sqrt{15}))\]

For more detailed explanation, please go to watch our video:


Published: 2023-05-09 22:09:49
Updated: 2023-05-09 22:53:02

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