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I have little problm to solve and if there is some complex analysis master, any help is appreciated.

I have to solve this equation.
$\sin^2(z) = i\pi$. Considering that my domain is the complex plane, there is no any kind of restrictions. I difference squares and in the and for the first solution I got in my equation as a part this expression. $\sqrt{1-i\pi}$. I know that there is a way to get rid of square root by adding some elements to this equation, but i can't remember how. I don't want to have any complex expression under the root, i.e. if I want to solve this equation I need to split real part of my expression and imaginary. How could I do that with this root?

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  • $\begingroup$ Have you considered a half-angle formula? $\endgroup$
    – abiessu
    Commented Dec 3, 2015 at 0:10
  • $\begingroup$ Note that $\sqrt{i} = (1+i)/2$. $\endgroup$
    – marty cohen
    Commented Dec 3, 2015 at 0:51

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You might want to use the connection between the complex cosine-function and the complex exponential function. $$\cos 2z = \frac{e^{2iz}+e^{-2iz}}{2}$$ and use the formula $$\cos 2z = 1 - 2\sin^2z\ .$$ Introducing $w = e^{2iz}$ the equation $\sin^2z = i\pi$ becomes a polynomial equation in $w\ .$ $$w^2 - 2w(1-2i\pi) + 1 = 0\ .$$

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  • $\begingroup$ Thank you. I made cardinal mistake, working with sin(z) and square roots of (iπ). This makes everything easier. Thank you very much. $\endgroup$
    – Ahmo
    Commented Dec 3, 2015 at 0:36
  • $\begingroup$ You are most welcome. I'm happy to have helped. $\endgroup$
    – Anders Muszta
    Commented Dec 3, 2015 at 1:20

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