Suppose you want to solve $\displaystyle \cos x + i \sin x = \text{cosh}(y-1) + ixy $. Without resorting to graphs could we do the following: $\displaystyle e^{ix} = \frac{e^{y-1}+e^{1-y}}{2} + ixy $?
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Originally Posted by manjohn12 Suppose you want to solve $\displaystyle \cos x + i \sin x = \text{cosh}(y-1) + ixy $. Without resorting to graphs could we do the following: $\displaystyle e^{ix} = \frac{e^{y-1}+e^{1-y}}{2} + ixy $? Yes But I don't think it's possible to solve it easily oO And maybe it's better to identify real and imaginary parts : cosh(y-1) is a real number. So cosh(y-1)=cos(x) and xy=sin(x)
So we have two cases: $\displaystyle x=0, \ x \neq 0 $.
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