How to solve for $x$ in $\sin(x)=i$

Question

So I was bored, and decided to do some math for fun. After a while, I came up with this:$$\text{Solve for }x\text{: }\sin(x)=i$$which I thought that I might be able to do.

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Here is my attempt at solving $\sin(x)=i$:

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So how do we even start to solve this equation? Well, we do it by expanding $\sin(x)$:$$\sin(x):=\dfrac12ie^{-ix}-\dfrac12ie^{ix}$$Now, we can solve for $\sin(x)=i$:$$\sin(x)=i$$$$\implies\dfrac12ie^{-ix}-\dfrac12ie^{ix}=i$$$$\implies\dfrac12e^{-ix}-\dfrac12e^{ix}=1$$$$\implies e^{-ix}-e^{ix}=2$$ $$e^{ix}\left(e^{-ix}-e^{ix}\right)=e^{ix}(2)$$$$1-e^{2ix}=2e^{ix}$$$$\implies2e^{ix}+e^{2ix}=1$$$$2e^{ix}+e^{2ix}=1$$$$\implies\dfrac12e^{ix}+1=\dfrac1{2e^{ix}}$$$$\implies e^{ix}+2=\dfrac1{e^{ix}}$$$$e^{2ix}+2e^{ix}-1=0$$$$a^2+2a-1=0$$Now, plugging this into the quadratic formula:$$e^{ix}=\dfrac{-2\pm\sqrt{4+4}}2$$$$e^{ix}=\dfrac{-\require{cancel}\cancel2^1\pm\cancel2^1\sqrt2}{\cancel2^1}$$$$e^{ix}=-1\pm\sqrt2$$However, for some weird reason, I had a feeling that the answer would be found in a hyperbolic trigonometric function (as I have never attempted solving this before), and as such, I decided to cancel out the positive outcome, so$$e^{ix}=-1-\sqrt2$$$$\implies ix=i\pi+2i\pi k+\ln(1+\sqrt2)$$$$\implies x=\pi+2\pi k-i\ln(1+\sqrt2),k\in\mathbb Z$$Or, we could write this as$$\pi+2\pi k-i\operatorname{arcsinh}(1)$$which can also be written as (more as so I don't have to write out \operatorname{arcsinh} or any other hyperbolic trigonometric function to the power of $-1$ with \operatorname,$$x=\pi+2\pi k-i\sinh^{-1}(1),k\in\mathbb Z$$

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My question

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Is my solution correct, or what could I do to attain the correct solution/attain it more quickly?

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Mistakes I might have made

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1. In all honesty, it might have not been a good idea to make the assumptions that I made throughout my finding the solution, but that's just me.
2. Remembering what $\sinh^{-1}(1)$ is equal to.
3. The most simplistic multiplication/division, since that always seems to trip me up for some reason.

Answer 1

As an alternative, we have that $x$ must be complex then by $x=u+iv$

$$\sin(x)=\sin(u+iv)=\sin u \cosh v+i\cos u \sinh v=i$$

which requires

• $\sin u=0 \implies u=2k\pi\;\lor \; u=\pi+2k\pi$

and then

• for $u=2k\pi \implies \cos u=1$ then $\sinh v=1 \implies v=\sinh^{-1}(1)$

• for $u=\pi+2k\pi \implies \cos u=-1$ then $\sinh v=1 \implies v=\sinh^{-1}(-1)$

that is

$$x=2k\pi +i\sinh^{-1}(1)\;\lor \; x=\pi+2k\pi +i\sinh^{-1}(-1)$$

which corresponds to your solution.