We are given with:
[tex]{\quad \qquad \longrightarrow \sin (z)={\sf a}\:,\:z\in \mathbb{C}}[/tex]
Recall the identity what we have for the sine function of complex numbers
- [tex]{\boxed{\bf{\sin (z)=\dfrac{e^{\iota z}-e^{-\iota z}}{2\iota}}}}[/tex]
Put the values to thus obtain:
[tex]{:\implies \quad \sf \dfrac{e^{\iota z}-e^{-\iota z}}{2\iota}=a}[/tex]
[tex]{:\implies \quad \sf e^{\iota z}-e^{-\iota z}=2a\iota}[/tex]
Multiply both sides by [tex]{\sf e^{\iota z}}[/tex]
[tex]{:\implies \quad \sf e^{\iota z}\cdot e^{\iota z}-e^{-\iota z}\cdot e^{\iota z}=2a\iota e^{\iota z}}[/tex]
[tex]{:\implies \quad \sf (e^{\iota z})^{2}-2a\iota e^{\iota z}-1=0}[/tex]
Put x = [tex]{\sf e^{\iota z}}[/tex]:
[tex]{:\implies \quad \sf x^{2}-2a\iota x-1=0}[/tex]
Find the discriminant, here D will be, D = (-2ai)² - 4 × 1 × (-1) = 4 - 4a² = 4(1-a²)
Now, By quadratic formula:
[tex]{:\implies \quad \sf x=\dfrac{-(-2a\iota)\pm \sqrt{4(1-a^{2})}}{2}}[/tex]
[tex]{:\implies \quad \sf x=\dfrac{a\iota \pm \sqrt{1-a^{2}}}{2}}[/tex]
[tex]{:\implies \quad \sf e^{\iota z}=\dfrac{a\iota \pm \sqrt{1-a^{2}}}{2}}[/tex]
[tex]{:\implies \quad \sf \iota z=log\bigg(\dfrac{a\iota \pm \sqrt{1-a^{2}}}{2}\bigg)}[/tex]
Using the formula for logarithms, we have:
[tex]{:\implies \quad \sf \iota z=log(a\iota \pm \sqrt{1-a^{2}})-log(2)}[/tex]
[tex]{:\implies \quad \sf z=\dfrac{1}{\iota}log(a\iota \pm \sqrt{1-a^{2}})-\dfrac{1}{\iota}log(2)}[/tex]
The sine function is periodic on 2πn and zero on (π/2), and the logarithmic expression becomes undefined for all ia±√(1-a²) < 0, so we will take modulus of it
[tex]{:\implies \quad \boxed{\bf{z=\dfrac{1}{\iota}log\bigg|a\iota \pm \sqrt{1-a^{2}}\bigg|-\dfrac{1}{\iota}log(2)+\dfrac{\pi}{2}+2\pi n\:\:\forall \:n\in \mathbb{Z}}}}[/tex]
