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

[am4th]

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.

 

[pines-demon]

$$\cos(z)=\cos(x+i y)=\cos(x)\cos(iy)-\sin(x)\sin(iy)=\cos(x)\cosh(y)-i\sin(x)\sinh(y)$$
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.

 

[FactChecker]

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

This is not right.

so cos(z) is bounded.

This would be true if the first statement was right, but it is not.

 

[vela]

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.

 

[martinbn]

Cosine is analytic in the whole plane, if it were bounded it would be constant by Liouville.

 

[PeroK]

$$\cos(z)=\cos(x+i y)=\cos(x)\cos(iy)-\sin(x)\sin(iy)=\cos(x)\cosh(y)-i\sin(x)\sinh(y)$$
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:
$$\cos(iy) = \cosh(y)$$

 

[mathwonk]

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.

 

[martinbn]

It also follows from Picard's little theorem.

 

[pasmith]

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.