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I am self-studying Complex Analysis and I am suppose to find $z\in \mathbb{C}$ such that $\sin z=100.$ I know that $$\sin z=\sin x \cosh y+i\cos x\sinh y$$

So I must have $\sin x \cosh y=100.$ I looked in Wolfram and I found that $y=i(x-\sin^{-1}(100))$. I was not able to solve this question. How does one find that solution? If fact, I was not expecting to find $y$ in function of $x$.

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You're looking at the wrong term first. Your goal is $$ 100 = \sin x \cosh y + i \cos x\sinh y$$ The most fruitful thing we can do here is to look at the imaginary part of each side which gives us $0=\cos x\sinh y$. A product can only be zero if at least one of its factors are, so we have either $\cos x = 0$ or $\sinh y = 0$. The latter possibility is true only for $y=0$ -- when $y=0$ the sine of $x+iy$ is indeed real, but it is also between $-1$ and $1$, so we won't find a solution there.

So we must have $\cos x=0$ which implies $\sin x = \pm 1$. Then look at the real part which now collapses into $100 = \pm \cosh y$ which you can now solve for $y$. It turns out that $\pm$ must be $+$ and $y$ must be $\pm\cosh^{-1}(100)$.

Now put your $x$ and $y$ together to find $$ z = \frac{\pi}2 + 2\pi k \pm i\cosh^{-1}(100) \qquad k\in\mathbb Z $$ The hyperbolic arccosine happens to be expressible in terms of simpler functions, so we can also write $$ z = \frac{\pi}2 + 2\pi k \pm i\log\left(100+\sqrt{9999}\right) \qquad k\in\mathbb Z $$

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In light of this answer, I think that Alpha returned a "$y$ in terms of $x$" answer because it was letting both $x$ and $y$ be complex. Restricting them to the reals is what gives the discrete set of solutions. Also, for the reader's awareness: the inverse for cosh and sinh follow from the hint I posted (it is standard). –  anon Apr 4 '12 at 2:02
    
@Henning Makholm: I found $z=\dfrac{\pi}{2}+k\pi+i\log(100\pm\sqrt{9999}),\,\,\,\,k\in\mathbb{Z}$. –  spohreis Apr 4 '12 at 21:21
    
@spohreis: It happens that $\log(100-\sqrt{9999})=-\log(100+\sqrt{9999})$ so you can put the $\pm$ either inside or outside the log (though I think outside is nicer because that makes it clear that the solutions come in conjugate pairs). However you need to have $2k\pi$ instead of $\pi$. With your formula, odd $k$s will give $\sin z=-100$ instead of $\sin z=100$. –  Henning Makholm Apr 4 '12 at 21:30
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Rewrite the sine function using complex exponentials,

$$\sin z = \frac{e^{iz}-e^{-iz}}{2i}.$$

If you then multiply both sides by $e^{iz}$ you'll have a quadratic equation in the variable $w=e^{iz}$...

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