Solving arcsin(√2) with Hyperbolic Sin Function

[Curious3141]

Your reasoning for $a=3\pi/2$ is flawed. When dealing with complex numbers, you can, in fact, take the log of a negative number, so you can't use that as a reason for ruling out that solution.

I'll probably regret butting in when I haven't read every single post (as I said, I find the method used unduly convoluted and tedious), but aren't $a$ and $b$ defined a priori to be real?

 

[vela]

Oh, yeah. Never mind.

 

[srfriggen]

Your reasoning for $a=3\pi/2$ is flawed. When dealing with complex numbers, you can, in fact, take the log of a negative number, so you can't use that as a reason for ruling out that solution.

I was going to say the other way I can see showing it is flawed is by inspecting e^b+e^-b and seeing that if b is assumed to be real it cannot equal -2$\sqrt{2}$.

Is case closed on this one Vela?

 

[vela]

Yup!

 

[srfriggen]

Yup!

Thank you so much Vela (and all contributors on this post) for all of your help, support, patience, and steering me in the right direction while still making me work for the answer. I definitely learned a lot and feel really good that I understand the problem and I appreciate the help!

 

[srfriggen]

I'm back with one more like this. I've worked through it very hard and I think there should be much less back and forth, as I really get what I'm doing now. But something in this problem isn't working... perhaps there is no solution.

sin^-1(2i)= a+bi, find a and b (and and b are real numbers).

sin(a+bi)=2i


(e^i(a+bi)-e^-i(a+bi))/2i = 2i

e^-b+ai-e^b-ai = 4i²

e^-be^ai-e^be^-ai=-4

e^-b[cos(a)+isin(a)] - e^b[cos(-a)+isin(-a)]=-4

e^-b[cos(a)+isin(a)] - e^b[cos(a)-isin(a)]=-4


{1} (e^-b-e^b)cos(a)=-4

{2} (e^-b+e^b)sin(a)=0


For {2}, a=0 and a=∏ are solutions.

Using a=0 in {1}: e^-b+e^b=-4

Solving this quadratic I get: b=ln(2±√5), which is not a Real solution when b=ln(2-√5).

Similarly, letting a=∏ (or 2∏ multiples of ∏) I get b=ln(-2±$\sqrt{5}$), which is not always real as well.

From what I see, letting a=n∏, where n is an integer, is the only plausible solution for equations {1} and {2}, yet they do not yield answers.

I reworked the problem twice and am still getting the same answers.

 

[vela]

I'll assume your {1} and {2} are correct. In solving, you flipped a sign. It should be $e^{-b}-e^b = \pm 4$. The exponentials are subtracted.

 

[Curious3141]

I'm back with one more like this. I've worked through it very hard and I think there should be much less back and forth, as I really get what I'm doing now. But something in this problem isn't working... perhaps there is no solution.

sin^-1(2i)= a+bi, find a and b (and and b are real numbers).

sin(a+bi)=2i(e^i(a+bi)-e^-i(a+bi))/2i = 2i

e^-b+ai-e^b-ai = 4i²

e^-be^ai-e^be^-ai=-4

e^-b[cos(a)+isin(a)] - e^b[cos(-a)+isin(-a)]=-4

e^-b[cos(a)+isin(a)] - e^b[cos(a)-isin(a)]=-4{1} (e^-b-e^b)cos(a)=-4

{2} (e^-b+e^b)sin(a)=0For {2}, a=0 and a=∏ are solutions.

Using a=0 in {1}: e^-b+e^b=-4

Solving this quadratic I get: b=ln(2±√5), which is not a Real solution when b=ln(2-√5).

Similarly, letting a=∏ (or 2∏ multiples of ∏) I get b=ln(-2±$\sqrt{5}$), which is not always real as well.

From what I see, letting a=n∏, where n is an integer, is the only plausible solution for equations {1} and {2}, yet they do not yield answers.

I reworked the problem twice and am still getting the same answers.

Perhaps I could suggest that solution I was thinking of (and which I find easier). Don't introduce exponentials (yet). Leave things in hyperbolic form.

Following the method in my post, you will end up with:

$\sin{a}\cosh{b} = 0$ [eq 1]

$\cos{a}\sinh{b} = 2$ [eq 2]

From 1, you should be able to see that $\cosh b$ can never equal zero. So $\sin a = 0$, yielding $a = n\pi$.

Put that into 2:

$\cos{n\pi}\sinh{b} = 2$

Consider separately $n$ odd and even, yielding:

$\sinh{b} = 2$ for even $n$

and $\sinh{b} = -2$ for odd $n$.

At this point, you can replace the $\sinh$ function with the exponential notation and solve the quadratic, etc. to get the value for b. Alternatively, just use the logarithmic formulation for the inverse hyperbolic sine directly (if you're allowed to) to get $arcsinh{2} = \ln(2 + \sqrt{5})$.

Either way, you should get the general solution quite easily (you can use $n = 2k$ and $n = 2k+1$ to express the even and odd cases more succinctly).

 

[srfriggen]

I'll assume your {1} and {2} are correct. In solving, you flipped a sign. It should be $e^{-b}-e^b = \pm 4$. The exponentials are subtracted.

I'm not sure what you mean by flipped a sign.

Do you mean when I was solving the quadratic? Or before that step?

 

[srfriggen]

Perhaps I could suggest that solution I was thinking of (and which I find easier). Don't introduce exponentials (yet). Leave things in hyperbolic form.

Following the method in my post 4595007, you will end up with:

$\sin{a}\cosh{b} = 0$ [eq 1]

$\cos{a}\sinh{b} = 2$ [eq 2]

From 1, you should be able to see that $\cosh b$ can never equal zero. So $\sin a = 0$, yielding $a = n\pi$.

Put that into 2:

$\cos{n\pi}\sinh{b} = 2$

Consider separately $n$ odd and even, yielding:

$\sinh{b} = 2$ for even $n$

and $\sinh{b} = -2$ for odd $n$.

At this point, you can replace the $\sinh$ function with the exponential notation and solve the quadratic, etc. to get the value for b. Alternatively, just use the logarithmic formulation for the inverse hyperbolic sine directly (if you're allowed to) to get $arcsinh{2} = \ln(2 + \sqrt{5})$.

Either way, you should get the general solution quite easily (you can use $n = 2k$ and $n = 2k+1$ to express the even and odd cases more succinctly).

I do see what you mean, and really appreciate this insight, but at this point I've written out the problem to be handed in very neatly and I'd rather just correct whatever mistake I have then re-work it.

 

[Curious3141]

I do see what you mean, and really appreciate this insight, but at this point I've written out the problem to be handed in very neatly and I'd rather just correct whatever mistake I have then re-work it.

No worries. It comes to pretty much the same idea as yours, but I just find it neater (and simpler) to leave the computation of $b$ to the end.

Basically, you've defined $b$ to be real. When you solve the quadratic for the respective cases, just take the solution that gives you a real logarithm (and hence a real $b$). Reject the other solution. It's that simple.

 

[srfriggen]

I'm not sure what you mean by flipped a sign.

Do you mean when I was solving the quadratic? Or before that step?

when solving the quadratic for a=0 this is what I get:

4±$\sqrt{(-4^2)-4(1)(-1)/2}$

= 4±$\sqrt{20}$/2

=2±√5 = e^b


b=ln(2±√5)

 

[Curious3141]

when solving the quadratic for a=0 this is what I get:

4±$\sqrt{(-4^2)-4(1)(-1)/2}$

= 4±$\sqrt{20}$/2

=2±√5 = e^b


b=ln(2±√5)

This is correct, but as I said in the edit in my last post, only $b = \ln (2 + \sqrt{5})$ is admissible, because $b$ is by definition real.

You should explicitly write that line into your derivation at the top, i.e. "Let the solution be $z = a + bi$, where $a$ and $b$ are real." EVERY complex number can be represented in this form, so there is no loss of generality.

 

[srfriggen]

This is correct, but as I said in the edit in my last post, only $b = \ln (2 + \sqrt{5})$ is admissible, because $b$ is by definition real.

You should explicitly write that line into your derivation at the top, i.e. "Let the solution be $z = a + bi$, where $a$ and $b$ are real." EVERY complex number can be represented in this form, so there is no loss of generality.

cool, I understand now. I just wasn't sure if I was "allowed" to dismiss the ln(2-√5).

but now I have values for a and b, namely, a=2pi+2npi and b=ln(2+√5).

Thanks!

 

[srfriggen]

cool, I understand now. I just wasn't sure if I was "allowed" to dismiss the ln(2-√5).

but now I have values for a and b, namely, a=2pi+2npi and b=ln(2+√5).

Thanks!

actually wait one second... if I choose a=∏ , then my b=ln(-2±√5), which is a real solution for ln(-2+√5).

So do I have two different a and b combination solutions? i.e. if a=pi, then b=n(2+√5), and if a=0, b=(2+√5)?

 

[Curious3141]

actually wait one second... if I choose a=∏ , then my b=ln(-2±√5), which is a real solution for ln(-2+√5).

So do I have two different a and b combination solutions? i.e. if a=pi, then b=n(2+√5), and if a=0, b=(2+√5)?

You should get:

$a = 2k\pi, b = \ln(\sqrt{5} + 2)$

and

$a = (2k+1)\pi, b = \ln(\sqrt{5} - 2)$

for integral k.

as your general solution. The first solution set corresponds to even $n$, while the second, to odd $n$. As I said before, consider the cases where $n$ is even and odd separately when solving for $b$. It is not sufficient to consider only the special cases where $a = 0$ or $a = \pi$ unless the question explicitly gives you bounds for $a$.