I How to see that cos(z) is unbounded, but cos(xy)+isin(xy) is bounded?

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Cos(z) is unbounded in the complex plane due to the behavior of its components, specifically that cosh(y) grows without bound as y increases, while cos(x) remains between -1 and 1. In contrast, the function f(x+iy) = cos(xy) + i sin(xy) is bounded because both cos(xy) and sin(xy) are limited to values between -1 and 1. The key distinction lies in the fact that while cos(z) is bounded on the real axis, it is not bounded across the entire complex plane, as established by Liouville's theorem. Additionally, the relationship between the boundedness of cos(z) and sin(z) indicates that if one is bounded, the other must be as well, which is not the case here. Therefore, cos(z) is indeed unbounded in the complex domain.
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I'm looking at a tutorial right now and using Liouville's theorem cos(z) is unbounded. On the next slide we are looking at the function f(x+iy)=cos(xy)+isin(xy), and the reasoning is that: |cos(xy)|≤1, |sin(xy)|≤1 so |f(x+iy)|≤2, so f(x+iy) is bounded.

I don't understand why this doesn't also apply to cos(z), that is: |cos(z)|≤1, so cos(z) is bounded.

Grateful for any insights.
 
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$$\cos(z)=\cos(x+i y)=\cos(x)\cos(iy)-\sin(x)\sin(iy)=\cos(x)\cosh(y)-i\sin(x)\sinh(y)$$
Now the real part is multiplied by ##-1\leq\cos(x) \leq 1## and ##\cosh(y)## which is unbounded above (times -1 is also unbounded below). Similar analysis works for the imaginary part.
 
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am4th said:
I don't understand why this doesn't also apply to cos(z), that is: |cos(z)|≤1,
This is not right.
am4th said:
so cos(z) is bounded.
This would be true if the first statement was right, but it is not.
 
am4th said:
I don't understand why this doesn't also apply to cos(z), that is: |cos(z)|≤1, so cos(z) is bounded.
The difference is that ##x## and ##y## are real whereas ##z## is complex. To put it another way, ##\cos z## is bounded on the real axis, but it's not bounded over the entire complex plane.
 
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Cosine is analytic in the whole plane, if it were bounded it would be constant by Liouville.
 
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pines-demon said:
$$\cos(z)=\cos(x+i y)=\cos(x)\cos(iy)-\sin(x)\sin(iy)=\cos(x)\cosh(y)-i\sin(x)\sinh(y)$$
Now the real part is multiplied by ##-1\leq\cos(x) \leq 1## and ##\cosh(y)## which is unbounded above (times -1 is also unbounded below). Similar analysis works for the imaginary part.
Or, let ##z = iy##, and note that:
$$\cos(iy) = \cosh(y)$$
 
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another argument, is to note that since cos^2 + sin^2 = 1, hence cos(z) is bounded if and only if sin(z) is bounded. And since e^(iz) = cos(z) + i.sin(z), hence if cos(z) is bounded then e^z is bounded. But e^z = e^(x+iy) = e^x.(e^iy) = e^x.(cos(y)+i.sin(y)), thus |e^z| =|e^x|, and I suppose you know that e^x is unbounded. E.g. e^x = 1+x + x^2/2 +.... > x, for x positive.
 
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It also follows from Picard's little theorem.
 
am4th said:
I'm looking at a tutorial right now and using Liouville's theorem cos(z) is unbounded. On the next slide we are looking at the function f(x+iy)=cos(xy)+isin(xy), and the reasoning is that: |cos(xy)|≤1, |sin(xy)|≤1 so |f(x+iy)|≤2, so f(x+iy) is bounded.

I don't understand why this doesn't also apply to cos(z), that is: |cos(z)|≤1, so cos(z) is bounded.

Grateful for any insights.

Note that if z = x + iy then
xy = \operatorname{Re}(z)\operatorname{Im}(z) = \frac{(z + \bar{z})(z - \bar{z})}{4i} so f(x + iy) = \cos(xy) + i\sin(xy) is not complex analytic.
 
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