I worked on a brain teaser that started with $\sin{(z)}=10.$ It then solved for $\cos{(z)}=\sqrt{-99}.$
I tried solving for $z,$ using $z=x+iy$ then $z=ae^{bi}$ but got nowhere. So is this even possible?
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- 1$\begingroup$ Yes. In fact, the only entire complex functions that are bounded are the constants. (Here, entire = analytic (in the complex sense) everywhere on $\mathbb{C}$. The $\sin$ function is entire.) $\endgroup$anomaly– anomaly2020-09-29 14:11:32 +00:00Commented Sep 29, 2020 at 14:11
- 1$\begingroup$ It's true that this shows that the (modulus of) the values of sine can be large, but it does not show that the sin itself ever takes on the value $10$. After all, $f(z) = \exp(z) + 10$ is unbounded, but it never takes on the value $10$.So it's something more subtle than being entire that makes this possible for sine. $\endgroup$John Hughes– John Hughes2020-09-29 14:26:43 +00:00Commented Sep 29, 2020 at 14:26
- 2$\begingroup$ This might be helpful $\endgroup$DatBoi– DatBoi2020-09-29 14:30:25 +00:00Commented Sep 29, 2020 at 14:30
- $\begingroup$ @JohnHughes: Yeah, that was more in the response to the question in the title rather than the one in the OP itself. $\endgroup$anomaly– anomaly2020-09-29 14:44:50 +00:00Commented Sep 29, 2020 at 14:44
- 1$\begingroup$ @JohnHughes: No: en.wikipedia.org/wiki/Picard_theorem . $\endgroup$anomaly– anomaly2020-09-29 14:52:12 +00:00Commented Sep 29, 2020 at 14:52
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2 Answers
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4 Yes.
For a complex number $z$, we have \begin{align} \sin z &= \frac{1}{2i}(\exp(iz) - \exp(-iz)) \\ \cos z &= \frac{1}{2}(\exp(iz) + \exp(-iz)) \end{align} so in your problem, we have \begin{align} 10 &= \frac{1}{2i}(\exp(iz) - \exp(-iz)) \\ 10i &= \frac{1}{2}(\exp(iz) - \exp(-iz)) & \text{mult through by $i$}\\ i \sqrt{99} &= \frac{1}{2}(\exp(iz) + \exp(-iz)) \end{align} Adding the last two equations gives us $$ i(10 + \sqrt{99} = \exp{iz} $$ so $$ \log i(10 + \sqrt{99} ) = iz $$ and hence $$ z = \frac{\log (i(10 + \sqrt{99}) ) }{i}. $$
Here "log" should be taken as the principal value of the natural logarithm (you may need to look this up!), and there's a second answer where we set the cosine to be $-i\sqrt{99}$, but I didn't bother writing it out.
- $\begingroup$ Thanks John and @datboi $\endgroup$Daniel– Daniel2020-09-29 17:12:09 +00:00Commented Sep 29, 2020 at 17:12
- 1$\begingroup$ The $\sin z$ formula is written incorrectly here which leads to an incorrect result. The denominator should be $2i$ instead of $2$. $\endgroup$Vladimir Fokow– Vladimir Fokow2021-12-09 19:19:12 +00:00Commented Dec 9, 2021 at 19:19
- 1$\begingroup$ Thanks Vladmir. Re-doing the math means that 10 in the last equation becomes 10i, as in your answer above. $\endgroup$Daniel– Daniel2021-12-09 22:51:19 +00:00Commented Dec 9, 2021 at 22:51
- $\begingroup$ Good catch; now fixed (I think). $\endgroup$John Hughes– John Hughes2021-12-10 22:42:29 +00:00Commented Dec 10, 2021 at 22:42
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Yes, it can.
Let us find such $z$ so that $\sin(z) = 10$.
Firstly, as a consequence of the Euler's formula: $ e^{iz} = \cos(z) + i \cdot \sin(z)$,
we have the following two identities which hold for all complex numbers $z$:
$$ \sin(z) = \frac{e^{iz} - e^{-iz}}{2i} \\ \cos(z) = \frac{e^{iz} + e^{-iz}}{2} $$
Then:
$$ \sin(z) = 10 \\ \frac{e^{iz} - e^{-iz}}{2i} = 10 \\ e^{iz} - e^{-iz} - 20i = 0 \\ (e^{iz})^2 - 20i\cdot(e^{iz}) - 1 = 0 \\ $$
Solving the quadratic equation:
$$ \sqrt{D}=\sqrt{(-20i)^2 - 4 \cdot 1 \cdot (-1)} = \sqrt{-396} = 6i\sqrt{11} \\ (e^{iz}) = \frac{-(-20i) \pm \sqrt{D}}{2} = i \cdot (10 \pm 3\sqrt{11}) \\ $$
Complex logarithm:
$$ iz = Ln\left(i \cdot \left(10 \pm 3\sqrt{11}\right)\right) = \ln\left(i \cdot \left(10 \pm 3\sqrt{11}\right)\right) + i \cdot 2\pi k, \qquad k \in \mathbb{Z}\\ iz = \ln(i) + \ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\ iz = \ln\left(e^{i\frac{\pi}{2}}\right) + \ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\ iz = i\frac{\pi}{2} + \ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\ $$
The result:
$$ z = \frac{\pi}{2} + 2\pi k - i \cdot \ln\left(10 \pm 3\sqrt{11}\right), \qquad k \in \mathbb{Z} $$
$\cos(z) = \sqrt{-99}$ can be solved in exactly the same way, giving the result:
$$ z = \frac{\pi}{2} + 2\pi k - i \cdot \ln\left(3\sqrt{11} \pm 10\right), \qquad k \in \mathbb{Z} $$
The idea is taken from this video: https://www.youtube.com/watch?v=3C_XD_cCeeI
