The sine inverse of a purely complex number

[Suvadip]

To prove that
$$sin^{-1}(ix)=2n\pi\pm i log(\sqrt{1+x^2}+x)$$
I can prove $$sin^{-1}(ix)=2n\pi+ i log(\sqrt{1+x^2}+x)$$

but facing problem to prove
$$sin^{-1}(ix)=2n\pi- i log(\sqrt{1+x^2}+x)$$

Help please

 

[alyafey22]

Re: Complex no

To prove that
$$sin^{-1}(ix)=2n\pi\pm i log(\sqrt{1+x^2}+x)$$
I can prove $$sin^{-1}(ix)=2n\pi+ i log(\sqrt{1+x^2}+x)$$

but facing problem to prove
$$sin^{-1}(ix)=2n\pi- i log(\sqrt{1+x^2}+x)$$

Help please

Can you sketch your proof ? what is $n$ ?

 

[mathbalarka]

Hint : $e^{ix} = \cos(x) + i\sin(x)$.

what is n?

An arbitrary integer of course.

 

[Suvadip]

Re: Complex no

Can you sketch your proof ? what is $n$ ?

n is any integer

Let $$sin^{-1}(ix)=\theta$$

Then $$sin\theta = ix$$

$$cos\theta=\pm\sqrt{1-sin^2\theta}=\pm\sqrt{1+x^2}$$

Now $$e^{i\theta}=cos\theta +i sin\theta=\pm\sqrt{1+x^2}+i i x=-x\pm\sqrt{1+x^2}$$

Therefore $$i\theta=Log (-x+\sqrt{1+x^2})(Only + sign ~~considered)=2n\pi i +log(-x+\sqrt{1+x^2})$$

$$\theta = 2n\pi-i log(-x+\sqrt{1+x^2})$$
$$\theta = 2n\pi+i log(x+\sqrt{1+x^2})$$

 

[mathbalarka]

In order to get the other sign, simply note that $\sin(-x) = -\sin(x)$.

 

[Suvadip]

In order to get the other sign, simply note that $\sin(-x) = -\sin(x)$.

Please a bit detailed hints