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I need to find the range of $h(z) = \frac{1}{z}$ if the domain is given by $0 < |z| \leq 1$ and $\frac{\pi}{2} \leq \operatorname*{Arg} z \leq \pi$.

$\\$ I know that if $w = \frac{1}{z}$, then $|w| \geq 1$ (since $0 < |z| \leq 1$). However, what I'm having a hard time figuring out is the argument's restriction. I know that $\operatorname*{Arg} w = \operatorname*{Arg} \frac{1}{z} = - \operatorname*{Arg} z$, but does that mean $ -\frac{\pi}{2} \geq \operatorname*{Arg} w \geq -\pi$?

I know that if the answer to the question is yes, then the range of $h(z)$ is the intersection of $|w| \geq 1$ and $-\frac{\pi}{2} \geq \operatorname*{Arg} w \geq -\pi$.

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    Well, I feel done with posting answers. My response is "yes, your reasoning is correct". – 2017-01-28
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    Thank you. I just wasn't sure if $- \frac{\pi}{2} \geq$ Arg w $\geq - \pi$ made sense. I'm not exactly sure how the argument can fit within that restriction. To me it's saying all quadrants except for the third one. – 2017-01-28
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    You could use the fact that $\arg(w)$ is in modulus $2\pi$ to make it positive if you wanted. – 2017-01-28
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    So is it the same as saying, $-\frac{\pi}{2} \leq$ Arg w $\leq \pi$? – 2017-01-28
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    Change both sides by adding $2\pi$. – 2017-01-28
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    Thanks. I think I'll just stick with the negative restrictions. Because adding 2$\pi$ tells me that Arg w is restricted to the third quadrant, and I'm not sure that makes sense to me. – 2017-01-28
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    If $\text{Arg}(w)\in [-\pi,-\pi/2]$, then $w$ lies in the third quadrant. – 2017-01-28
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    The restriction states that it's smaller than $-\frac{\pi}{2}$ and greater than $-\pi$. That wouldn't be in the third quadrant. – 2017-01-28

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