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Math Help - How to solve sin(z)=2 (Complex Analysis)

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    How to solve sin(z)=2 (Complex Analysis)

    The problem wants me to solve sin(z)=2 by equating the real and imaginary parts.

    Using sin(z)=sin(x+iy), the definition of sin(z), and some algebra, I've gotten as far as:
    e^{xi}e^{-y}-e^ye^{-ix}=4i.
    This is where I'm stuck. I know I'll have to set the imaginary part to 4 and the real part to 0, but how do I separate them from this point...?
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    Quote Originally Posted by paupsers View Post
    The problem wants me to solve sin(z)=2 by equating the real and imaginary parts.

    Using sin(z)=sin(x+iy), the definition of sin(z), and some algebra, I've gotten as far as:
    e^{xi}e^{-y}-e^ye^{-ix}=4i.
    This is where I'm stuck. I know I'll have to set the imaginary part to 4 and the real part to 0, but how do I separate them from this point...?
    \sin(z)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)
    So let \sin(x)\cosh(y)=0 and \cos(x)\sinh(y)=4.
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