am4th
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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.
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.