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Find the domain of f(z)=arcsin(abs(z+i)).

I know that the domain of arcsin(x) is between -1 and 1, but how do you find the domain when the inside is complex?

MatheMat
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  • because of the absolute value, the inside is real and in fact non-negative. So, you just have to figure out when is $|z+i|\in [-1,1]$, which is equivalent to asking when is $|z+i|\leq 1$? Think geometrically :) – peek-a-boo Apr 19 '21 at 05:21
  • How do you find when abs(z+i) less than equal to 1. I've just begun learning complex numbers so I'm not too sure how to approach this. – MatheMat Apr 19 '21 at 05:29
  • Think about a pictorial representation: in general for $a\in\Bbb{C}$ and $r\geq 0$, what shape is described by the inequality $|z-a|\leq r$? In words, this is saying that $z$ is at most a distance $r$ away from the point $a$. I'm sure elementary geometry will tell you what shape to consider (if it is still not obvious, ask yourself, what is the region described by $|z|\leq 1$). – peek-a-boo Apr 19 '21 at 05:30
  • So I've had a think and I'm pretty sure it's a circle.abs(z+i) is less than or equal to 1 is a circle shaded in with radius 1. How does that relate to the domain or arcsin(abs(z+i))? – MatheMat Apr 19 '21 at 05:38
  • Yes, so the domain of $z\mapsto \arcsin(|z+i|)$ is precisely the closed disc of radius $1$ centered at $-i$. That's all there is to it. – peek-a-boo Apr 19 '21 at 05:39
  • Thanks for the help, do you also know how to find the implied range of the function? – MatheMat Apr 19 '21 at 05:42
  • how about you let me know what you have tried, then I'll give you a hint. – peek-a-boo Apr 19 '21 at 05:43
  • the range of arcsin(x) is [-pi/2,pi/2] for the original. Since there's no vertical shifts to the function, would the range remain the same? – MatheMat Apr 19 '21 at 05:45
  • Ok, the range of $\arcsin$ is $[-\pi/2,\pi/2]$, but that is only if you allow $x$ to vary over the entire interval $[-1,1]$. In your case, $x=|z+i|$, so what are the values $x$ can take? Therefore, what is the corresponding image? – peek-a-boo Apr 19 '21 at 05:49
  • in re How do you prove the domain of a function?, i guess what's meant here is find all the values such that $f$ is defined? – BCLC Apr 19 '21 at 06:40

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