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How would I show that $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$?

Im not sure how to begin, does it involve using $\sinh z=\frac{e^{z}-e^{-z}}{2}$ and $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$?

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Yes, it does. Also, that $|u+iv|^2=u^2+v^2$. –  Cameron Buie May 26 '12 at 0:10
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As a general rule, if you have an idea, try it. Only for the simplest of problems (and problems like those you have previously solved) will you know the right thing to do before you do it. If your idea works and you solve the problem, great! If it doesn't work, then when you come here, you'll be able to get much deeper advice if you show what you've tried and why you think it doesn't work. –  Hurkyl May 26 '12 at 0:24
    
Thanks, I did try some calculations using pen and paper, but I didnt really get anywhere with it and I was not sure if I was on the right direction, so hence I didnt type out what I've written out. Seeing what Javier Badia just answered below, I did actually use the sum formula for $\sin(a+b)$, but I just didnt realise to could convert it as he did below. –  Derrick May 26 '12 at 0:27
    
@Derrick: Even if you aren't sure you're on the right track, it's a good idea to show what you've done so far. People are going to be more inclined to give you detailed and complete answers if they can see that you've made an effort. In this case, you were on the right track. You had all the relevant pieces, and it was just a matter of figuring out how they were strung together. That's very reassuring to see. –  Cameron Buie May 26 '12 at 0:38
    
@CameronBuie and @ Hurkyl , ok noted, will try to make my questions more constructive in the future. –  Derrick May 26 '12 at 0:48
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3 Answers 3

If you use the sine addition formula, the pythagorean identities, and the fact that $\sin(ix)=i\sinh (x)$ and $\cos(ix) = \cosh(x)$, then you get this:

$$ \begin{align} \sin(x+iy) &= \sin x \cos (iy)+\cos x \sin(iy) \\ &= \sin x \cosh y + i \cos x \sinh y \end{align} $$

$$ \begin{align} |\sin(x+iy)|^2 &= (\sin x \cosh y)^2 + (\cos x \sinh y)^2 \end{align} $$

Now you can get rid of the cosines knowing that $\cos^2 x + \sin^2 x = 1$ and that $\cosh^2 x - \sinh ^2 x = 1$. You can take it from there.

By the way, to get the sine addition formula and the sine and cosine of imaginary numbers, convert them to exponential form:

$$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$$ $$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$ $$\sinh x = \frac{e^x-e^{-x}}{2}$$ $$\cosh x = \frac{e^x+e^{-x}}{2}$$

Plug in what you want to find out; the derivation of the identities is straightforward.

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Thanks @JavierBadia , I just have one question, Im not sure how you would expand out $(\sin x \cosh y)^2 + (\cos x \sinh y)^2$? What would the next line be? I get $\sin^2x\cosh^2y +\cos^2x\sinh^2y$? –  Derrick May 26 '12 at 0:43
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@Derrick: the idea is that we want to be left only with the circular and hyperbolic sines, so we use the identities I mentioned to replace the cosines with sines. I'm on my iPod so I can't really type out all the TeX, but that is the gist of it. –  Javier Badia May 26 '12 at 0:56
    
Okay thanks, most apreciated. Its ok, I think I got it from here. –  Derrick May 26 '12 at 0:58
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$$z=x+iy\Longrightarrow \sin z=\frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{-y+ix}-e^{y-ix}}{2i}=$$$$=\frac{e^{-y}(\cos x+i\sin x)-e^y(\cos x-i\sin x)}{2i}=\frac{1}{2i}\left[i\sin x\left(e^y+e^{-y}\right)-\cos x\left(e^y-e^{-y}\right)\right]=$$$$=\sin x\cosh y+i\cos x\sinh y\Longrightarrow ...$$

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\begin{align} \sin(z)^2 &= (\sin x \cos (iy))^2 +(\cos x \sin(iy))^2 \\ &=\sin^2x \cosh^2y+\cos^2x \sinh^2y \\ &= \sin^2x (1+\sinh^2y )+(1-\sin^2x ) \sinh^2y \\ &=\sin^2 x+\sin^2 x \sinh^2y+ \sinh^2y-\sin^2x \sinh^2y \\ &=\sin^2x+\sinh^2y \end{align}

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